We call a set of points in the affine plane in general position if no $3$ of them are collinear. Let $a_n(q)$ be the number of (ordered) $n$-tuples of pairwise points in $\mathbb{F}_q^2$ in general position.
By the comments,
\begin{align*}
a_0(q) & = 1 \\
a_1(q) & = q^2 \\
a_2(q) & = q^2(q^2 - 1) & & = (q+1)q^2(q-1) \\
a_3(q) & = q^2(q^2 - 1)(q^2 - q) & & = (q+1)q^3(q-1)^2 \\
a_4(q) & = q^2(q^2 - 1)(q^2 - q)(q^2 - 3q + 3) & & = (q+1)q^3(q-1)^2(q^2 - 3q + 3)
\end{align*}
The goal of this answer is to determine $a_5(q)$.
For a set $S$ of points, we call a line of $\mathbb F_q^2$ determined by $S$ if it contains at least two points of $S$.
A set $S$ is in general position determines $\binom{\#S}{2}$ lines, as no line can contain more than $2$ points of $S$.
There are $q^2(q^2-1)(q^2-q)$ triples $T$ of points in general position. They are all isomorphic, so for the further extendability it is enough to investigate $T = ((0,0),(1,0),(0,1))$.
We study the different cases $x$ for the extensions $T' = T \cup \{x\}$.
To get a $4$-tuple $T'$ in general position, $x$ must be one of the $q^2 - (3 + 3(q-2)) = q^2 - 3q + 3$ points covered by the three lines determined by $T$.
The points $y$ such that $T'' = T' \cup\{y\}$ is a $5$-tuple of distinct points in general position are precisely the ones not contained in the $6$ lines determined by $T'$. So if the $6$ lines spanned by the point pairs in $T\cup\{x\}$ cover $e$ points, the number of subsequent extensions by a point $y$ is $q^2 - e$. Unlike the last step, the number $e$ depends on the choice of the point $x$.
Look at the set of $\mathcal{L}$ of six lines determined by $T'$.
Two distinct lines of the form $\overline{ab}, \overline{ac}\in\mathcal L$ have $a$ as their intersection point, which is in $T'$.
There are three line pairs of the form $\{\overline{ab},\overline{cd}\}$ with $T = \{a,b,c,d\}$. The intersection point (if it exists) of such a line pair is not contained in $T'$. Also, no two intersection points of this type can be identical (we would have an $a\in T'$ such that two lines $L,L'$ in $\mathcal{L}$ pass through $a$ and a point outside of $T'$. Hence $L = L'$, which contradicts the property of $T'$ being in general position). Hence the number $3-f$ of intersection points outside of $T'$ of the lines in $\mathcal L$ (and in turn the number $e$) depends on the number $f \in\{0,1,2,3\}$ of parallel line pairs in $\mathcal{L}$. We get that $e = 4 + 6(q-2) - (3-f) = 6q -11 + f$.
As we will see below, the number of parallel line pairs $f$ determined by a quadruple of points behaves differently in even and odd characteristic.
For each $a\in T$, there is a unique line $L_a$ passing through $a$ which is parallel to the line spanned by the other two points in $T$.
We have $L_{(0,0)} = \langle (1,-1)\rangle$, $L_{(1,0)} = (1,0) + \langle (0,1)\rangle$ and $L_{(0,1)} = (0,1) + \langle (1,0)\rangle$.
The points arising as the intersection of two of these lines are $(1,1)$, $(1,-1)$ and $(-1,1)$.
Case 1: $q$ is odd.
- If $x$ is one of these three points, $f = 2$.
- If $x$ is one of $3(q-1) - 3\cdot 2 = 3q - 9$ points on exactly one of the lines $L_{\ast}$, $f = 1$.
- If $x$ is one of the remaining $q^2 - 3q + 3 - 3 - (3q - 9) = q^2 -6q + 9$ points ouside of $T$, $f = 0$.
In total, we get $a_5(q) = q^2 (q^2 - 1)(q^2 - q)\Big(3(q^2 - 6q + 9) + (3q - 9)(q^2 - 6q + 10) + (q^2 - 9q + 15)(q^2 - 6q + 11)\Big) = q^2(q^2-1)(q^2-q)(q^4 - 9q^3 + 32q^2 - 54q + 36)$.
Case 2: $q$ is even
- Here, the three listed points $(1,1)$, $(1,-1)$ and $(-1,1)$ are one and the same. For this single point $x$, $f = 3$.
- If $x$ is one of the $3(q-2) = 3q - 6$ points on a single line $L_{\ast}$, $f = 1$.
- If $x$ is one of the remaining $q^2 - 3q + 3 - 1 - (3q - 6) = q^2 -6q + 8$ points outside of $T$, $f = 0$.
In total, we get $a_5(q) = q^2 (q^2 - 1)(q^2 - q)\Big(1\cdot(q^2-6q+8) + (3q-6)(q^2 - 6q + 10) + (q^2 - 6q + 8)(q^2 - 6q + 11)\Big) = q^2 (q^2 - 1)(q^2 - q)(q^4 - 9q^3 + 32q^2 - 54q + 36)$
Conclusion
To my surprise, the expressions in both cases coincide and hence
$$a_5(q) = (q+1)q^3 (q-1)^2(q-2)(q-3)(q^2 -4q + 6)$$
is still a polynomial expression in $q$.
Remark 1
It might be easier to investigate this question in the projective plane over $\mathbb F_q$ instead of the affine plane $\mathbb F_q^2$.
Then any two lines are intersecting, so no analysis of parallel lines is needed. It is pretty straightforward to come up with the numbers
\begin{align*}
b_0(q) & = 1, \\
b_1(q) & = q^2 + q + 1, \\
b_2(q) & = (q^2 + q + 1)(q^2 + q) & & = (q^2 + q + 1)(q+1)q\\
b_3(q) & = (q^2 + q + 1)(q^2 + q)q^2 & & = (q^2 + q + 1)(q+1)q^3 \\
b_4(q) & = (q^2 + q + 1)(q^2 + q)q^2(q^2-2q + 1) & & = (q^2 + q + 1)(q+1)q^3(q-1)^2
\end{align*}
The determination of $b_5$ can be done similarly as of $a_5$ above. It is easier since (unlike in the affine situation) the collineation group of the projective plane still acts transitively on the set of $4$-tuples of points in general position. It leads to
$$b_5(q) = b_4(q) \cdot ((q^2+q+1)-(4 + 6(q-1) - 3)) = (q^2+q+1)(q+1)q^3(q-1)^2(q-2)(q-3),$$
and in that case it is clear that no $q$ even/odd distinction is necessary. Maybe there is a way to derive $a_5$ from $b_5$, resolving the "surprise" of the two identical expressions we got above.
I expect the computation of higher $b_n$'s to be more complicated, as I think it will involve a case by case analysis also in the projective case (at least as long as noone told us a general counting principle...).
Remark 2
The presence of the factors $(q-2)$ and $(q-3)$ in the computed expressions for $a_5$ and $b_5$ is a good sign, as in the affine and projective planes over $\mathbb F_2$ and $\mathbb F_3$, it is clear that there is no set of $5$ points in general position:
The largest size of a set of points in general position (both in the affine and projective plane) is a classical topic in finite geometry and known as the maximal size of a $2$-arc. For $q$ even, the largest $2$-arcs are the hyperovals of size $q+2$, and for $q$ odd, the largest $2$-arcs are the ovals of size $q+1$.
