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 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 $\mathbf{Z}^5$.