Yes. This is given by OEIS sequence A160870. The number of subgroups of index $n$ in $\mathbf{Z}^m$ is there denoted $T(n,m)$. There is a closed formula in terms of the divisors of $n$ given at this page. Recursively, we have: $$ T(m,1) = 1, \quad T(1,n) = 1, \quad T(m,n) = \sum_{d \mid m} \left(\frac{m}{d}\right)^{n-1} \cdot T(d, n-1) $$
For example, there are $T(5,4) = 651$ subgroups of index $4$ in $\mathbf{Z}^5$.
