There is no upper bound for finite index subgroups of $SL_n({\mathbb Z})$. That is, given any integer $k$, there exists a finite index subgroup of $SL_n({\mathbb Z})$ which needs at least $k$ generators. I do not know to whom this is originally due, but it is a remark of Rapinchuk that if you pull the centre of, say $SL_{2n}({\mathbb Z}/m {\mathbb Z})$ where $m$ is the product of $k$ distinct primes back to $SL_{2n}({\mathbb Z}) $, you get a finite index subgroup of $SL_{2n}({\mathbb Z})$ whose abelianisation has $({\mathbb Z}/2{\mathbb Z})^k$ as a quotient and hence is at least $k$ generated.
On the other hand, a finite index subgroup of $SL_n({\mathbb Z})$, contains a smaller finite index subgroup which is generated by 3 elements; this result is true for any higher rank non-uniform arithmetic group in a semi-simple linear group (and is due to Ritumoni Sarma and Venkataramana)