This is just a brief answer. The only autmorphisms other than the ones that you know about already arise from elements of $H^1(H,V)$ (with $H = {\rm SL}(n,q)$ or ${\rm GL}(n,q)$), where the corresponding automorphisms induce the identity on $V$ and on $G/V \cong H$, but map a complement $H$ of $V$ to a complement that is not conjugate to $H$ in $G$.

These cohomology groups are all known. The only cases in which they are non-zero are when $H = {\rm SL}(2,2^k)$ with $k \ge 2$, and $H = {\rm SL}(3,2) = {\rm GL}(3,2)$, and in each of these cases, the dimension of $H^1(H,V)$ is $1$-dimensional over ${\mathbb F}_q$.

So, for example, for $G = {\rm ASL}(2,8)$, the order of the automorphism group is $24|{\rm AGL}(2,8)| = 168|G|$, with a factor $3$ coming from the field automorphisms of ${\mathbb F}_8$, and the factor $8$ coming from $H^1(H,V)$.