(For simplicity I will discuss $G = SL_n$. Making the necessary modifications to $GL_n$ shouldn't be hard.)
The most important result here is Steinberg's restriction theorem, which tells you that the restriction of $L(\lambda)$ to $G(\mathbb{F_q})$ is simple if $\lambda$ is of the form $\lambda = \sum \lambda_i \omega_i$ with $0 \le \lambda_i \le q-1$ for $1 \le i \le n$ ($\omega_i$ are the fundamental weights). (Let us call this set of dominant weights $X_q$).
The second most important result here is Steinberg's tensor product theorem, which tells you that for any highest weight $\lambda$ if we write
$ \lambda = \lambda_0 + p \lambda_1 + \dots + p^a \lambda_a$
with each $\lambda_i \in X_p$ then $L(\lambda) = L(\lambda_0) \otimes L(\lambda_1)^{Fr} \otimes \dots \otimes L(\lambda_a)^{Fr^a}$ (where $M^{Fr}$ denotes the Frobenius twist of a module $M$, i.e. inflation with respect to the Frobenius $Fr : G \to G$.) This allows you to reduce to the case $q = p$.
Hence one needs to understand the modules $L(\lambda)$ for $\lambda$ in the box $X_p$ decompose. As you mention, these are the simple socles of induced modules $\nabla(\lambda)$.
For $n = 2$ this is easy: each $L(\lambda) = \nabla(\lambda)$ for $0 \le \lambda \le p-1$. Hence $L(\lambda)$ is realised as homogenous polynomials in $k[X,Y]$ of degree $\lambda$ (with natural $SL_2$ action).
For $n = 3$ things get more complicated. (Here the relevant $\nabla(\lambda)$ can have either 1 or 2 composition factors, and one gets the answer using the Jantzen sum formula. (See Jantzen's "Representations of algebraic groups").
For $n = 4$ things get even more complicated! Here I think one can still get the answer from Jantzen's sum formula, but this seems more of an accident here.
In general there is a character formula (due to Lusztig) which holds for large $p$ (see Jantzen's book). In the case of $n = 4$ this is known to hold for any $p > n$.
