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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 recursive formula in terms of the divisors of $n$ given at this page. The initial conditions are: $$ T(n,1) = 1, \quad T(1,m) = 1 $$ and recursively, we have either: $$ \quad T(n,m) = \sum_{d \mid n} \left(\frac{n}{d}\right)^{m-1} \cdot T(d, m-1) $$ or $$ \quad T(n,m) = \sum_{d \mid n} d \cdot T(d, m-1) $$ For example, there are $T(5,4) = 651$ subgroups of index $4$ in $\mathbf{Z}^5$.