$\newcommand\ZZ{\mathbb Z}\newcommand\cl{\overline{\mathbb F_p}}$This solution will show that the claimed answer, i.e. $n/6$ when $6\mid n$, $\lceil n/4\rceil$ when $6\nmid n$ but $2\mid n$, and $\lceil n/3\rceil$ when $2\nmid n$, is correct for all $n$. The proof is based on classification of sets of four vertices of regular polygons which lie together in some lattice. The main claim is that any four such vertices either form a rectangle centered at the center of the $n$-gon (and $2\mid n$) or comprise four of the six vertices of a regular hexagon centered at the center of the $n$-gon (and $6\mid n$).
To prove this, we first need to transfer the problem into algebraic language.
Initial claim. Label the vertices of a regular $n$-gon $P_0P_1\cdots P_{n-1}$ in counterclockwise order, and let $\Phi_n$ be the $n$th cyclotomic polynomial. Then the points $P_a$, $P_b$, $P_c$, and $P_d$ lie on some lattice if and only if there exist $r,s,t,u\in\ZZ$, all nonzero, with $r+s+t+u=0$ and $\gcd(r,s,t,u)=1$, for which
$$\Phi_n(x) \mid rx^a+sx^b+tx^c+ux^d.$$
Proof. Let $\zeta=e^{2\pi i/n}$, and situate the regular $n$-gon in the complex plane such that $P_k=\zeta^k$ for each $0\leq k<n$. The four points lie on a lattice if and only if they lie on a lattice with $P_a\mapsto 0$, i.e. if and only if $P_b-P_a$, $P_c-P_a$, and $P_d-P_a$ are linearly dependent over $\ZZ$. This happens if and only if there exist $s,t,u\in\ZZ$, not all zero, with
$$s(\zeta^b-\zeta^a)+t(\zeta^c-\zeta^a)+u(\zeta^d-\zeta^a)=0.$$
Such an identity holds if and only if it holds for all Galois conjugates of $\zeta$; since $\Phi_n$ has no repeated roots, this is the same as $\Phi_n(x)\mid rx^a+sx^b+tx^c+ux^d$, with $r=-s-t-u$. Note that it is not possible for any of $\{r,s,t,u\}$ to be zero without all being zero; if two are zero (say $t=u=0$) then $r=s\neq 0$ and $\zeta^a=\zeta^b\implies P_a=P_b$, and if only one is zero (say $u$) then $P_a,P_b,P_c$ must be collinear. $\square$
Part 1. We show the result for all $n=2^e3^f$. This section is mostly technicalities, and doesn't contain that many interesting ideas.
We first treat the case where $n=2^e$ is a power of $2$; here $\Phi_n(x)=x^{n/2}+1$. Say
$$x^{n/2}+1\ \big|\ rx^a+sx^b+tx^c+ux^d$$
and consider taking the result of the right side modulo the left side (as a polynomial with degree $<n/2$). If one of $\{a,b,c,d\}$ (say, $a$) has a unique residue modulo $n/2$, this will result in a term $rx^a$ (if $a<n/2$) or $-rx^{a-n/2}$ (if $a\geq n/2$) which cannot be cancelled, giving a contradiction. Since at most two terms can agree modulo $n/2$, this means that the terms must exist in pairs modulo $n/2$, i.e. (without loss of generality) $a\equiv c\pmod{n/2}$ and $b\equiv d\pmod{n/2}$. This means that $P_a$ and $P_c$ are diametrically opposite vertices, as are $P_b$ and $P_d$, and $P_aP_bP_cP_d$ is a rectangle.
We now treat the case where $n=3^f$ is a power of $3$; here $\Phi_n(x)=x^{2n/3}+x^{n/3}+1$. Say
$$x^{2n/3}+x^{n/3}+1\ \big|\ rx^a+sx^b+tx^c+ux^d$$
and consider taking the result of the right side modulo the left side (as a polynomial with degree $<2n/3$). Similarly to before, if there are at most $2$ exponents in $\{a,b,c,d\}$ in a particular residue class modulo $n/3$, it is impossible for them to cancel. So, three of the exponents $\{a,b,c,d\}$ must agree modulo $n/3$, but then the fourth has to be alone in its residue class modulo $n/3$, as there are only three elements of each residue class modulo $n/3$ in $[0,n)$, a contradiction.
Finally, we treat the case where $n=2^e3^f$ and $e,f>0$; here $\Phi_n(x)=x^{n/3}-x^{n/6}+1$. Say
$$x^{n/3}-x^{n/6}+1\ \big|\ rx^a+sx^b+tx^c+ux^d$$
and consider taking the result of the right side modulo the left side (as a polynomial with degree $<n/3$). Again, if only one of $\{a,b,c,d\}$ lies in a particular residue class modulo $n/6$, this term can never be cancelled, so either all four of $\{a,b,c,d\}$ lie in the same residue class modulo $n/6$ or two fall in each of two residue classes. In the first case, $\{P_a,P_b,P_c,P_d\}$ form four vertices of a hexagon, and in the second (say $a\equiv c$ and $b\equiv d\pmod{n/6}$) we must have
$$y^2-y+1\mid ry^{a'}+ty^{c'},\ sy^{b'}+uy^{d'}$$
with $y=x^{n/6}$ and $a'=\left\lfloor \frac{a}{n/6}\right\rfloor$, et cetera. Since $r\zeta^{a'}=-t\zeta^{c'}$ when $\zeta^2-\zeta+1=0$, so for some $\zeta$ with $|\zeta|=1$, we must have $|r|=|t|$, and so $y^2-y+1$ must divide $y^{a'}\pm y^{c'}$; this implies $3\mid a'-c'$, and so $3\mid b'-d'$. This means that $a\equiv c$ and $b\equiv d\pmod{n/2}$, and so $P_aP_bP_cP_d$ is a rectangle, as desired.
Part 2. We treat the case where other primes divide $n$. The main result of this part will be the following.
Proposition. If $P_a,P_b,P_c,P_d$ are vertices of a regular $n$-gon lying on a lattice, $p\mid n$ is a prime greater than $3$, and $e=\nu_p(n)$, then either $a\equiv b\equiv c\equiv d\pmod{p^e}$ or $P_aP_bP_cP_d$ forms a rectangle.
We show this mainly by consideration of when $\Phi_n(x)$ divides polynomials with few terms and coefficients summing to $0$ in $\mathbb F_p[x]$. We first need a rather technical lemma about binomial coefficients, for which Lucas's theorem will be crucial.
Lemma 1. Let $p$ be a prime and $e$ be a positive integer, and consider for $0\leq a<p^e$ the vectors
$$v_{a,e}=\left(\binom ak\bmod p\right)_{0\leq k<p^{e-1}(p-1)}\in \cl^{p^{e-1}(p-1)}$$
and
$$w_{a,e}=\left(\binom ak\bmod p\right)_{0\leq k<p^e}\in \cl^{p^e}$$
The $w_{a,e}$ are linearly independent, and any $p-1$ of the $v_{a,e}$ are linearly independent.
Proof. We first show the result for $e=1$. Here, the $i$th entry of $v_{a,e}$ is a degree $i$ polynomial in $a$, and so, taking linear combinations of entries,
$$\sum_{a=0}^{p-1}\lambda_av_{a,1}=0\implies \sum_{a=0}^{p-1}\lambda_a a^k=0\text{ for all }0\leq k<p-1$$
and
$$\sum_{a=0}^{p-1}\mu_aw_{a,1}=0\implies \sum_{a=0}^{p-1}\mu_a a^k=0\text{ for all }0\leq k\leq p-1$$
If one of the $\lambda_a$ is zero, the vector of remaining $\lambda_a$ is an eigenvector of a $(p-1)\times (p-1)$ Vandermonde matrix with zero eigenvalue, and so the other $\lambda_a$ must be zero as well. The $\mu_a$ must form an eigenvector of a $p\times p$ Vandermonde matrix with zero eigenvalue, and so they must all be zero.
Now, we show the result for $e>1$. For an integer $q$, let $a\bmod q$ denote the remainder when $a$ is divided by $q$, and for $0\leq i<e$ let $f_i$ be the function mapping $a$ to the $p^i$s digit of $a$ in base $p$. Then, by Lucas's theorem,
$$\binom ak\equiv\binom{\sum_{i=0}^{e-1}p^if_i(a)}{\sum_{i=0}^{e-1}p^if_i(k)}\equiv\prod_{i=0}^{e-1}\binom{f_i(a)}{f_i(k)}\pmod p,$$
so
$$w_{a,e}=w_{f_0(a),1}\otimes w_{f_1(a),1}\otimes \cdots \otimes w_{f_{e-1}(a),1},$$
where $\otimes$ denotes a tensor product over $\cl$. This means that the set of $w_{a,e}$ is the $e$-fold tensor product of the set of $w_{a,1}$ for $0\leq a<p$, and so they are all linearly independent. Now, using Lucas's theorem again,
$$v_{a,e}=w_{a\bmod p^{e-1},e-1}\otimes v_{\lfloor a/p^{e-1}\rfloor,1}.$$
Since all of the distinct vectors $w_{a\bmod p^{e-1},e-1}$ are linearly independent,
$$\sum_a\lambda_a v_{a,e}=0\implies \sum_{a\bmod p^{e-1}=b}\lambda_av_{\lfloor a/p^{e-1}\rfloor,1}=0\text{ for all }0\leq b<p^{e-1}.$$
If the sum on the right is nonempty for some $b$, the property for $e=1$ implies that there are at least $p-1$ nonzero $\lambda_a$, as desired. $\square$
With this lemma, we can prove a somewhat general result about multiples of $\Phi_n$ in $\mathbb F_p[x]$ with few terms.
Lemma 2. Let $n$ be a positive integer and let $p$ be a prime dividing $n$. Let $n=p^em$ with $p\nmid m$. Any multiple of $\Phi_n$ in $\mathbb F_p[x]$ of degree less than $n$ either has at least $p$ nonzero terms or is of the form
$$\sum_{s=0}^{p^e-1}x^sq_s\left(x^{p^e}\right),$$
where the $q_s\in\mathbb F_p[x]$ are (not necessarily nonzero) multiples of $\Phi_m$ of degree less than $m$.
Proof. The identity
$$\Phi_n(x)=\frac{\Phi_m(x^{p^e})}{\Phi_m(x^{p^{e-1}})}$$
holds over $\ZZ$. Now, since $\Phi_m\in\mathbb F_p[x]$, the $(\cdot)^p$ distributes over the polynomial, and so
$$\Phi_n(x)=\frac{\Phi_m(x^{p^e})}{\Phi_m(x^{p^{e-1}})}=\frac{\Phi_m(x)^{p^e}}{\Phi_m(x)^{p^{e-1}}}=\Phi_m(x)^{p^{e-1}(p-1)}.$$
So, if $\Phi_n(x)$ divides some polynomial of degree less than $n$ with $L<p$ nonzero terms, then there exist distinct $0\leq a_1,\dots,a_L<n$ and $r_1,\dots,r_L\in\mathbb F_p$, not all zero, such that
$$\Phi_m(x)^{p^{e-1}(p-1)}\ \bigg|\ \sum_{i=1}^Lr_ix^{a_i}.$$
This means that, for each root $\zeta$ of $\Phi_m(x)$ in $\cl$, substituting $y=x-\zeta$, that
$$y^{p^{e-1}(p-1)}\ \bigg|\ \sum_{i=1}^Lr_i(y+\zeta)^{a_i}=\sum_{k=0}^ny^k\left(\zeta^{-k}\sum_{i=1}^Lr_i\zeta^{a_i}\binom{a_i}k\right).$$
In particular,
$$\sum_{i=1}^Lr_i\zeta^{a_i}\binom{a_i}k=0\text{ for all }0\leq k<p^{e-1}(p-1).$$
For $0\leq k<p^e$, $\binom ak$ is a function solely of $a\bmod p^e$, so we can let $S\subset\ZZ/p^e\ZZ$ be the set of residues of $\{a_1,\dots,a_\ell\}$ modulo $p^e$ (note $|S|\leq L$) and get
$$\sum_{s\in S}\binom sk\left(\sum_{a_i\bmod p^e=s}r_i\zeta^{a_i}\right)=0\text{ for all }0\leq k<p^{e-1}(p-1).$$
Since $|S|\leq L\leq p-1$, Lemma 1 implies that the inner sum is zero for all $s\in S$. Letting $b_i=\lfloor a_i/p^e\rfloor$ for each $i$, one now has
$$\sum_{i=1}^L r_ix^{a_i}=\sum_{s\in S}x^s\sum_{a_i\bmod p^e=s}r_i\left(x^{p^e}\right)^{b_i}=\sum_{s\in S}x^sq_s\left(x^{p^e}\right),$$
with $q_s(x)=\sum_{a_i\bmod p^e=s}x^{b_i}$. Now, $q_s(\zeta^{p^e})=0$ for every root $\zeta$ of $\Phi_m$; since $x\mapsto x^{p^e}$ permutes the roots of $\Phi_m$ and $\Phi_m$ has no repeated roots, this means $\Phi_m\mid q_s$ for each $s$, as desired. $\square$
We now need two results about multiples of some $\Phi_m$ with $\gcd(m,p)=1$ in $\mathbb F_p[x]$.
Lemma 3. Let $p$ be an odd prime, and let $\gcd(m,p)=1$. Suppose that $a$ is a positive integer and $c\in\cl$ such that $\zeta^a=c$ for all roots $\zeta$ of $\Phi_m$ in $\cl$. Then either $c=1$ and $m\mid a$ or $m$ is even, $c=-1$, and $\frac{2a}m$ is a positive odd integer.
Proof. Since $\zeta\mapsto\zeta^{-1}$ permutes the roots of $\Phi_m$, we must have
$$1=\zeta^a\zeta^{-a}=c^2,$$
so $c=\pm 1$. If $c=1$, then $a\mid m$, since $\Phi_m$ shares no roots with $x^k-1$ when $m\nmid k$ (any such roots would be roots of $x^{\gcd(k,m)}-1$ and thus double roots of $x^m-1$, which don't exist in $\cl$). If $c=-1$, then $\zeta^{2a}=1$ for every root $\zeta$ of $\Phi_m$, and so $m\mid 2a$ for similar reasons; however, $m$ cannot divide $a$, since otherwise $\zeta^a=1\neq -1$. $\square$
Lemma 4. Let $p$ be a prime number and let $m$ be a positive integer with $\gcd(m,p)=1$. Suppose that, as elements of $\mathbb F_p[x]$, $\Phi_m(x)\mid rx^a+sx^b+tx^c$ for distinct $0\leq a,b,c<m$ and $r,s,t\in\mathbb F_p$ with $r+s+t=0$. Then $r=s=t=0$.
Proof. Firstly, if one of $\{r,s,t\}$ is nonzero, we may without loss of generality let $s$ be nonzero, and scale $r$, $s$, and $t$ so that $s=-1$. Also, since factors of $x$ are irrelevant, we may replace the triple $a,b,c\in\ZZ/m\ZZ$ with $a-c,b-c,0\in\ZZ/m\ZZ$; in other words, we assume $c=0$. Now, we have $\Phi_m(x) \mid r(x^a-1)-(x^b-1)$, i.e. that $r(\zeta^a-1)+1=\zeta^b$ for every root $\zeta$ of $\Phi_m$. Since $\zeta^{-1}$ is also a root of $\Phi_m$ for all such $\zeta$, we have
$$1=\zeta^b\zeta^{-b}=\big(r(\zeta^a-1)+1\big)\big(r(\zeta^{-a}-1)+1\big)=(r^2-r)\big(\zeta^a-1\big)^2\zeta^{-a}$$
for every root $\zeta$ of $\Phi_m$, so either $r\in\{0,1\}$ or $\Phi_m(x)\mid x^a-1$. By Lemma 2, since $b<m$ and $b\neq c=0$, we cannot have $r=0$, as this implies $\Phi_m(x)\mid x^b-1$. Similarly, since $a<m$ and $a\neq c=0$, we cannot have $\Phi_m(x)\mid x^a-1$. However, $r=1$ implies $\Phi_m(x)\mid x^b-x^a$, which cannot happen since $m\mid b-a$. $\square$
We now are ready to prove the main proposition of this section. Suppose that $\Phi_n$ divides $rx^a+sx^b+tx^c+ux^d$ over $\mathbb F_p[x]$ with $r+s+t+u=0$. Note that we may assume at least one of $\{r,s,t,u\}$ is nonzero, since the setup over $\ZZ$ has $\gcd(r,s,t,u)=1$. Since $p>3$, Lemma 2 implies that
$$rx^a+sx^b+tx^c+ux^d=\sum_{i=0}^{p^e-1}x^iq_i\left(x^{p^e}\right),$$
where each $q_i\in\mathbb F_p[x]$ is a multiple of $\Phi_m$ of degree less than $m$. The number of nonzero terms in the polynomial (modulo $p$) is at most $4$, and it is also exactly the sum of the number of nonzero terms over all $q_i$. No $q_i$ may have exactly one nonzero term (since $\Phi_m\nmid x^a$ for all $a$). If some $q_i$ has exactly three nonzero terms, then either all other $q_j$ are zero, in which case the fact that $r+s+t+u=0$ means the existence of $q_i$ contradicts Lemma 4, or some other $q_j$ has exactly one term, a contradiction. So, either there is one $q_i$ with exactly four nonzero terms, or there are one or two $q_i$ with exactly two nonzero terms. In the first case, $a\equiv b\equiv c\equiv d\pmod{p^e}$, and in the second case Lemma 3 tells us that the only possible nonzero $q_i$ are products of $x^{m/2}+1$ and some monomial (since $\deg q_i<m$). If there is only one nonzero $q_i=\alpha x^\beta(x^{m/2}+1)$, the sum of the coefficients $r+s+t+u=0$ is exactly $2\alpha$, which implies $\alpha=0$, a contradiction. If there are two nonzero $q_i$, then two pairs of the exponents $\{a,b,c,d\}$ are equivalent modulo $m/2\cdot p^e=n/2$, and $P_aP_bP_cP_d$ forms a rectangle, as desired. $\square$
We now put everything together. Suppose that there exists some $n$ and four vertices $P_aP_bP_cP_d$ of a regular $n$-gon which lie on a lattice but neither form a rectangle nor comprise four of the six vertices of a regular hexagon; take the smallest such $n$ for which this is the case. By Part 1, $n$ must have some prime factor $p$ greater than $3$; let $n=p^em$ with $\gcd(m,p)=1$. By the main proposition of Part 2, since $P_aP_bP_cP_d$ do not form a rectangle, $a\equiv b\equiv c\equiv d\pmod{p^e}$. This means that $P_a$, $P_b$, $P_c$, and $P_d$ actually comprise four vertices of a regular $m$-gon, and so there are four vertices of a regular $m$-gon which form neither a rectangle nor a subset of vertices of a regular hexagon. This contradicts the minimality of $n$, and so we are done.
