I'm pretty sure that no such explicit generating set exists in the literature. However, at some point I spent some time thinking about this and I have a conjectural generating set.

**Normal generators.** Let $\Gamma_n$ denote $SL_n(\mathbb{Z})$ and let $\Gamma_n(p)$ denote the level $p$ congruence subgroup of $\Gamma_n$. For $1 \leq i,j \leq n$ such that $i \neq j$, let $e_{ij}$ be the $(i,j)$-elementary matrix. Define $S = \{e_{ij}\ |\ 1 \leq i,j \leq n, i \neq j\}$ and $S(p) = \{e_{ij}^p\ |\ 1 \leq i,j \leq n, i \neq j\}$. Clearly we have $S(p) \subset \Gamma_n(p)$. Moreover, the proof of the congruence subgroup theorem (by Mennicke and Serre) shows that $\Gamma_n(p)$ is the normal closure in $\Gamma_n$ of the subgroup generated by $S(p)$.

**Abelianization of congruence subgroup.**
However, $S(p)$ does not generate $\Gamma_n(p)$. This follows from the computation of the abelianization of $\Gamma_n(p)$ by Lee and Szczarba. Briefly, they defined a homomorphism $$\phi : \Gamma_n(p) \rightarrow \mathfrak{sl}_n(\mathbb{Z}/p).$$
Here $\mathfrak{sl}_n(\mathbb{Z}/p)$ is the abelian group of $n \times n$ matrices of trace $0$ with entries in $\mathbb{Z}/p$. The definition of $\phi$ is as follows. An element of $\Gamma_n(p)$ is of the form $1 + p A$ for some matrix $A$. Define $\phi(1+pA) = A$ modulo $p$. It is easy to see that the trace of $A$ mod $p$ is $0$. Moreover, this map is a homomorphism due to the identity
$$(1+pA)(1+pB) = 1+p(A+B)+p^2 AB.$$
Lee and Szczarba proved two things about $\phi$.

- They proved that $\phi$ is surjective. I'll say more about this below.
- They proved that the kernel of $\phi$ (which, by the way, is clearly $\Gamma_n(p^2)$) is the whole commutator subgroup of $\Gamma_n(p)$ for $n \geq 3$.

This implies that the abelianization of $\Gamma_n(p)$ is $\mathfrak{sl}_n(\mathbb{Z}/p)$.

**They only normally generate.**
Now, above I claimed that $S(p)$ does not generate $\Gamma_n(p)$. If $G$ is the subgroup generated by $S(p)$, it is easy to see that $\phi(G)$ is exactly the group of matrices whose diagonals are $0$. The problem thus is that you don't get the diagonal matrices.

**Hitting the diagonal.**
We thus need generators that hit the diagonal matrices. For $1 \leq i < n$, let $f_i$ be the result of inserting the $2 \times 2$ matrix $\left( \begin{array}{ll} 1+p & -p \\ p & 1-p\\ \end{array} \right)$ into the $n \times n$ identity matrix with the upper left at position $(i,i)$. Define $T = \{f_i\ |\ 1 \leq i < n\}$.
I conjecture that $S(p) \cup T$ generates $\Gamma_n(p)$. Here is my evidence for this.

Let $H$ be the subgroup generated by $S(p) \cup T$. Then $\phi(H)$ is all of $\mathfrak{sl}_n(\mathbb{Z}/p)$.

I can prove this by hand for $p=2$ and $p=3$.

**Conjectural proof sketch.**
My proof for $p=2$ and $p=3$ should work in general, so let me briefly describe it. By the congruence subgroup property business discussed above, the normal closure in $\Gamma_n$ of $H$ is $\Gamma_n(p)$. To prove that $H = \Gamma_n(p)$, therefore, it is enough to prove that $H$ is normal. To do this, it is enough prove the following. Let $e_{ij} \in S$ be an elementary matrix (a generator for $\Gamma_n$). Then for $s \in S(p) \cup T$, it is enough to prove that $e_{ij} s e_{ij}^{-1}$ and $e_{ij}^{-1} s e_{ij}$ can be written as words in $S(p) \cup T$. For $p=2$ and $p=3$, this is fairly easy, but I haven't managed to do the computations for higher $p$.