Yes. This is given by OEIS sequence [A160870](https://oeis.org/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:
$$ 
T(m,1) = 1, \quad T(1,n) = 1, \quad T(m,n) = T \left(\sum_{d \mid m} d \cdot T(d, n-1) \right)
$$

 For example, there are $T(5,4) = 651$ subgroups of index $4$ in $\mathbf{Z}^5$.