In a more general setting, let $A$ be a commutative ring, $I, J$ its ideals, $n\geqslant3$, then
$$[E(n,A,I),E(n,A,J)]\geqslant E(n,R,IJ),$$
where $E(n,A,I)$ is the normal closure in $E(n,A)$ of the subgroup $E(n,I)$ generated by the elementary generators $x_{ij}(\xi)=1+\xi e_{ij}$ of level $I$, that is, with $\xi\in I$. In particular, $SL(n,A)$ is perfect once it coincides with $E(n,R)$ (which is the case for polynomials over a field).

This is well-known since Bass' 1964 paper. Since $\mathbb{F}_q[T]$ is Euclidean (and thus of stable rank $2$), one has $E(n,R,I)=SL(n,R,I)$ by the $K_1$-stability result of Vaserstein (Theorem 3.2).

It is also known that $E(n,R,I)=\langle z_{ij}(\xi,\eta)\colon \xi\in I,\ \eta\in A \rangle$, where $z_{ij}(\xi,\eta)=x_{ji}(\eta)x_{ij}(\xi)x_{ji}(-\eta)$.

Now the projection $SL(n,A,I)\to SL(n,A,I)_{\mathrm{ab}}$ factors through $S=SL(n,R,I)/SL(n,r,I^2)$, and it is not hard to show for $A=\mathbb{F}_q[T]$ and $I=fA$ that the image of $z_{ij}(\xi,\eta)$ is $S$ has finite order, hence this abelianization is also finite.

Alternatively, one can do it with an even more straightforward computation. A reasonably good set of generators is known for $[E(n,A,I),E(n,A,J)]$ (see this paper). Namely, it is generated (as a normal subgroup) by the elements of the following three types:

- $[x_{ij}(\xi),z_{ij}(\zeta,\eta)]$,
- $[x_{ij}(\xi),x_{ji}(\zeta)]$,
- $x_{ij}(\xi\zeta)$,

where $\xi,\zeta\in I$, $\eta\in A$.

Now $z_{ij}(\xi)^k=x_{ji}(\eta)x_{ij}(k\xi)x_{ji}(-\eta) \equiv x_{ij}(k\xi)$, so taking $k=\operatorname{char}(\mathbb{F}_q)$ gives you the finiteness of orders of the generators, while the third relation shows that it suffices to take only finitely many of them and allows to calculate the abelianization explicitly (similar to this answer).