4 included Andy's remark

I don't seem to have seen any explicit generators for the principal congruence subgroups of $SL(n, \mathbb{Z}),$ for $n>2,$ although it is known (Sury+Venkataramana) is that the number of generators is bounded independently of the modulus. It is certainly possible to write a program to generate the generators (since the subgroup is given as a kernel of a map to a finite group, if we fix a generating set, the kernel is generated by the loops in the induced Cayley graph of the image group), but this seems painful (and will probably produce really awful generating sets without much further work).

For the self-proclaimed obtuse The question is: is there a known way to produce, given $n, p,$ a generating set of the level $p$ principal congruence subgroup in $SL(n, \mathbb{Z}).$ Particularly a nice, small generating set.

EDIT The Sury and Venkataramana paper is actually available for free. and having overcome my laziness somewhat, I see that they do actually construct a generating set using an idea somewhat related to @Andy's, but the generating set is quite large. Since $SL(n, \mathbb{Z})$ itself can be generated by three elements, one is tempted to make a rash conjecture that the minimal size of a generating set of $\Gamma_n(p)$ (with the obvious notation) is bounded independently of $n$ and $p.$ BUT (EDIT), as pointed out by @Andy Putman in his answer and his comment, this would be rash indeed, since the rank of the abelianization of $\Gamma_n(p)$ is $n^2-1.$

3 added more detail and commented on the references.

I don't seem to have seen any explicit generators for the principal congruence subgroups of $SL(n, \mathbb{Z}),$ for $n>2,$ although it is known (Sury+Venkataramana) is that the number of generators is bounded independently of the modulus. It is certainly possible to write a program to generate the generators (since the subgroup is given as a kernel of a map to a finite group, if we fix a generating set, the kernel is generated by the loops in the induced Cayley graph of the image group), but this seems painful (and will probably produce really awful generating sets without much further work).

For the self-proclaimed obtuse The question is: is there a known way to produce, given $n, p,$ a generating set of the level $p$ principal congruence subgroup in $SL(n, \mathbb{Z}).$ Particularly a nice, small generating set.

EDIT The Sury and Venkataramana paper is actually available for free. and having overcome my laziness somewhat, I see that they do actually construct a generating set using an idea somewhat related to @Andy's, but the generating set is quite large. Since $SL(n, \mathbb{Z})$ itself can be generated by three elements, one is tempted to make a rash conjecture that the minimal size of a generating set of $\Gamma_n(p)$ (with the obvious notation) is bounded independently of $n$ and $p.$

I don't seem to have seen any explicit generators for the principal congruence subgroups of $SL(n, \mathbb{Z}),$ for $n>2,$ although it is known (Sury+Venkataramana) is that the number of generators is bounded independently of the modulus. It is certainly possible to write a program to generate the generators (since the subgroup is given as a kernel of a map to a finite group, if we fix a generating set, the kernel is generated by the loops in the induced Cayley graph of the image group), but this seems painful (and will probably produce really awful generating sets without much further work).
For the self-proclaimed obtuse The question is: is there a known way to produce, given $n, p,$ a generating set of the level $p$ principal congruence subgroup in $SL(n, \mathbb{Z}).$ Particularly a nice, small generating set.