Sure, you can certainly work something out if you like. First of all note that $\textrm{GL}(n,\mathbb{Z}/m\mathbb{Z})$ is the product over all prime powers $p^a$ exactly dividing $m$ of $\textrm{GL}(n,\mathbb{Z}/p^a\mathbb{Z})$, so you're reduced to the case where $m$ is a prime power.

Second, there is an exact sequence $1 \to U_1 \to \textrm{GL}(n,\mathbb{Z}/p^a\mathbb{Z}) \to \textrm{GL}(n,\mathbb{Z}/p\mathbb{Z}) \to 1$ where I'll use $U_i$ to denote all mod $p^a$ matrices that are congruent to the identity modulo $p^i$. Given the Steinberg generators for $\textrm{GL}(n,\mathbb{Z}/p\mathbb{Z})$, you're reduced to giving generators/relations for $U_1$ (as well as working out how lifts of the Steinberg generators act by conjugation on the generators for $U_1$, as well as what elements of $U_1$ you get by applying the Steinberg relations to the lifts of your generators).

Note that $U_1$ is a big $p$-group, and each $U_i/U_{i+1}$ up to $i=a-1$ is just isomorphic to $\text{M}_n(\mathbb{Z}/p\mathbb{Z})$, so again you can produce generators and relations by devissage. In practice it shouldn't require so many generators as you get out of this. For instance for $n=2$, the group $U_1$ is generated by four elements: the upper-triangular unipotent matrix with $p$ in the upper-right, its transpose, and diagonal matrices with diagonal entries $c,1$ and $1,c$, where $c$ is a generator for the multiplicative group $1 + p\mathbb{Z}/p^a\mathbb{Z}$.

Sure, you can certainly work something out if you like. It may well be that an off-the-rack answer to your question is in the literature somewhere, but let me say how I would go about getting there from here.

Note that $\textrm{GL}(n,\mathbb{Z}/m\mathbb{Z})$ is the product over all prime powers $p^a$ exactly dividing $m$ of $\textrm{GL}(n,\mathbb{Z}/p^a\mathbb{Z})$, so you're reduced to the case where $m$ is a prime power.

Second, there is an exact sequence $1 \to U_1 \to \textrm{GL}(n,\mathbb{Z}/p^a\mathbb{Z}) \to \textrm{GL}(n,\mathbb{Z}/p\mathbb{Z}) \to 1$ where I'll use $U_i$ to denote all mod $p^a$ matrices that are congruent to the identity modulo $p^i$. Given the Steinberg generators for $\textrm{GL}(n,\mathbb{Z}/p\mathbb{Z})$, you're reduced to giving generators/relations for $U_1$ (as well as working out how lifts of the Steinberg generators act by conjugation on the generators for $U_1$, as well as what elements of $U_1$ you get by applying the Steinberg relations to the lifts of your generators).

Note that $U_1$ is a big $p$-group, and each $U_i/U_{i+1}$ up to $i=a-1$ is just isomorphic to $\text{M}_n(\mathbb{Z}/p\mathbb{Z})$, so again you can produce generators and relations by devissage. In practice it shouldn't require so many generators as you get out of this. For instance for $n=2$, the group $U_1$ is generated by four elements: the upper-triangular unipotent matrix with $p$ in the upper-right, its transpose, and diagonal matrices with diagonal entries $c,1$ and $1,c$, where $c$ is a generator for the multiplicative group $1 + p\mathbb{Z}/p^a\mathbb{Z}$.