Why is the quotient map $SL_n(\mathbb{Z})$ to $SL_n(\mathbb{Z}/p\mathbb Z)$ is surjective?

The result is true for $n\geq 1$ and any integer $q\geq 1$.

The group $SL_n(\mathbb{Z}/q\mathbb{Z})$ is generated by the elementary (transvection) matrices. It is easily seen that every elementary matrix is in the image of $\pi$, as the image of an elementary matrix in $SL_n(\mathbb{Z})$. So $\pi$ is surjective.

It suffices to prove that $SL_n(\mathbb{Z}/q)$ is generated by the elementary matrices (the matrices with ones along the diagonal and exactly one other nonzero entry). Equivalently, given any matrix in $SL_n(\mathbb{Z}/q)$, you can reduce it to $I$ using elementary row and column operations (add a multiple of a row to another row, or add a multiple of a column to another column).

Let $A\in SL_n(\mathbb{Z}/q)$, and consider its bottom row. By performing elementary column operations you can effectively perform the Euclidean algorithm, reducing the entries in the bottom row to all zero except for one entry which must be invertible modulo $q$. Moving this entry to the bottom-right position and making it a $1$, we reduce to the $n-1$ case.