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 $$ T(n,1) = 1 \quad \textrm{ for all } n\in \mathbb{N}$$ and recursively for $m > 1$, we have either $$ \quad T(n,m) = \sum_{d \mid n} \left(\frac{n}{d}\right)^{m-1} \cdot T(d, m-1) $$ or equivalently, $$ \quad T(n,m) = \sum_{d \mid n} d \cdot T(d, m-1) $$ Note that we can solve this recurrence to get the "explicit" formula $$ T(n,m) = \sum_{\substack{(d_0,d_1,\ldots,d_m)}} d_1 \cdots d_{m-1}$$ where the sum is over all sequences of integers $(d_0,d_1,\ldots,d_m)$ with $d_0=1$, $d_m=n$, and $d_i \mid d_{i+1}$ for all $i=0,\ldots,m-1$.
For example, there are $T(4,5) = 651$ subgroups of index $4$ in $\mathbf{Z}^5$.
