In general
$$F(\alpha) = \sum_{n=1}^\infty \frac{\cot(n\pi \alpha)}{n^3}$$
converges and can be explicitly calculated when $\alpha$ is a quadratic irrational. The convergence in this case is easily seen as $\alpha$ has irrationality measure $2$. More precisely, $F(\alpha)/\pi^3 \in \mathbb{Q}(\alpha)$ when $\alpha$ is quadratic irrational.
The procedure below also works when $n^3$ is replaced by any $n^{2k+1}$.
Let $$g(z) = \frac{\cot(z\pi \alpha)\cot(z\pi)}{z^3}$$ then $g$ has simple poles at non-zero integer multiples of $1$ and $1/\alpha$, and $5$-th order pole at $0$. Let $R_N$ denote a large rectangle with corners at $N(\pm 1 \pm i)$. Then contour integration gives
$$\tag{1}\sum_{\substack{n\in R_N \\ n\neq 0}} \frac{\cot(n\pi\alpha)}{\pi n^3} + \sum_{\substack{n/\alpha\in R_N \\ n\neq 0}} \frac{\alpha^2\cot(n\pi/\alpha)}{\pi n^3}-\frac{\pi ^2 \left(\alpha^4-5 \alpha^2+1\right)}{45 \alpha} = \frac{1}{2\pi i} \int_{R_N} g(z)dz$$
I claim there exists a sequence of integers $N_1, N_2, \cdots$ such that RHS tends to $0$. Note that $\cot(z\pi)$ is uniformly bounded on the annulus $R_{N+3/4} - R_{N+1/4}$ when $N$ is an integer. Hence by equidistribution of $n\alpha$ modulo $1$, we can find integers $N_i$ such that both $\cot(z\pi\alpha)$ and $\cot(z\pi)$ are uniformly bounded on $R_{N_i+3/4} - R_{N_i+1/4}$.
Since we already know the series converges, from $(1)$:
$$\tag{2}F(\alpha) + \alpha^2F(\frac{1}{\alpha}) = \underbrace{\frac{\pi ^3 \left(\alpha^4-5 \alpha^2+1\right)}{90 \alpha}}_{\rho(\alpha)}$$
Note that obviously $F(\alpha+1)=F(\alpha)$.
Let the continued fraction expansion of $\alpha$ be given by
$$\alpha = [a_0;a_1,a_2,\cdots]$$
Successive complete quotients are denoted by:
$$\zeta_0 = [a_0;a_1,a_2,\cdots]\qquad \zeta_1 = [a_1;a_2,a_3,\cdots]\qquad \zeta_2 = [a_2;a_3,a_4,\cdots]$$
Then $(2)$ and periodicity implies for $k\geq 0$:
$$\tag{3} F(\zeta_{k+1}) + \zeta_{k+1}^2 F(\zeta_k) = \rho(\zeta_{k+1})$$
If continued fraction of $\alpha$ is of form
$$\alpha = [a_0;a_1,\cdots,a_m,\overline{b_1,\cdots,b_r}]$$
Then $\zeta_{m+r+1} = \zeta_{m+1}$, so we eventually entered a cycle. $(3)$ gives a system of $m+r+1$ linear equations (by setting $k=0,\cdots,m+r$), with $m+r+1$ variables: $F(\zeta_0), F(\zeta_1),\cdots,F(\zeta_{m+r})$.
$$\begin{cases}
F(\zeta_1) + \zeta_1^2 F(\zeta_0) &= \rho(\zeta_1) \\
F(\zeta_2) + \zeta_2^2 F(\zeta_1) &= \rho(\zeta_2) \\
\cdots \\
F(\zeta_{m+1}) + \zeta_{m+1}^2 F(\zeta_{m+r}) &= \rho(\zeta_{m+1})
\end{cases}$$
Solving it gives the value of $F(\zeta_0)=F(\alpha)$.
A few examples: for $\alpha = (1+\sqrt{5})/2$, the continued fraction has period $1$, direct substitution into $(2)$ gives
$$\sum_{n=1}^\infty \frac{\cot(n\pi \frac{1+\sqrt{5}}{2})}{n^3} = -\frac{\pi ^3}{45 \sqrt{5}}$$
Complexity of result increases as period of $\alpha$ increases. For $\alpha = \sqrt{211}$, which has period $26$:
$$\sum_{n=1}^\infty \frac{\cot(n\pi \sqrt{211})}{n^3} = \frac{128833758679 \pi ^3}{383254107060 \sqrt{211}}$$
Another example:
$$\sum_{n=1}^\infty \frac{\cot(n\pi(\frac{1}{4}+\frac{\sqrt{7}}{3}))}{n^5} =
\frac{\left(3882947003 \sqrt{7}+14895286560\right) \pi ^5}{3344720117760}$$
The closed-form here follows immediately by noting $\csc x = \cot (x/2) - \cot x$.
Inspired by a (now deleted) answer of Si-Qi Liu, I wrote the following Mathematica code to evaluate such sums efficiently. $\sum_{n=1}^\infty \frac{\cot(n\pi \alpha)}{n^{2m+1}}$ means cotsum[alpha,m]. The command cotsum[Sqrt[61],1] evaluates the sum in the question.
cotsum[a0_?QuadraticIrrationalQ, m_Integer] /; m >= 1 :=
Block[{f, z, list, const, mat, resfunc, a},
resfunc[aa_] :=
Evaluate[Residue[Cot[z] Cot[aa z]/z^(2 m + 1), {z, 0}]];
a = ToNumberField[a0]; f[aa_] := f[aa - Floor[aa]] /; aa > 1;
list =
DeleteDuplicates@
NestWhileList[1/# - Floor[1/#] &, a - Floor[a], UnsameQ, All];
{const, mat} =
CoefficientArrays[f[#] + #^(2 m) f[1/#] == resfunc[#] & /@ list,
f /@ list];
Pi^(2 m + 1)/2*RootReduce[LinearSolve[mat, const][[1]]]];
