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 $H^1(H,V)$.
As a quick justification, note that $V$ is clearly characteristic in $G$, so ${\rm Aut}(G)$ induces a subgroup of the automorphism group of $G/V$. The full automorphism group of ${\rm SL}(n,q)$ is $\Gamma L(n,q)(.2)$, where the $.2$ is the duality automorphism when $n \ge 3$. The duality automorphism of $G/V$ does not lift to an automorphism of $G$, but we do have $A \Gamma L(n,q) \le {\rm Aut}(G)$.
Since $G$ acts absolutely irreducibly on $V$, automorphisms of $G$ inducing the idenity on $G/V$ must act as scalars on $V$, and we already have the full group of scalars present in $A \Gamma L(n,q)$.
So it remains to consider automorphisms of $G$ that induce the identity on both $G/V$ and on $V$ and, modulo inner automorphisms induced by conjugation by elements of $V$, these correspond exactly to $H^1(H,V)$.
A reference for the results on $H^1(H,V)$ is: G.W. Bell, On the Cohomology of the Finite Special Linear Groups, I, Journal of Algebra 54, 216-238 (1978), but I am not sure whether that was the first proof.