Not only these subgroups are finitely presented, they all finite products of cyclic groups; most of them (for sufficiently large $n$) are actually free of finite rank (once congruence subgroup contains no elements of order 2 and 3). Rank is easily computable if you know index of the congruence subgroup in the modular group. The magic formula is multiplicativity of the Euler characteristic: For the modular group $\Gamma$, $\chi=-1+\frac{1}{2} + \frac{1}{3}=-\frac{1}{6}$. If $\Gamma'\subset \Gamma$ is a subgroup of index $i$ then $\chi(\Gamma')=i \chi(\Gamma)$. If $\Gamma$ is free of rank $r$ then $\chi(\Gamma)= 1-r$. For instance, to find index $i$ for $\Gamma(n)$, compute the order of the quotient group $PSL(2, Z_n)$. There is a (somewhat ugly) closed formula for the order of this group (in terms of prime factors of $n$) which will tell you what the index is:
If $n$ is the product of powers of primes $\prod_i p_i^{k_i}$ then $$ |PSL(2,Z_n)|= \frac{1}{2} n^3 \prod_i (1- p_i^{2}). $$