David Treumann's guess is correct: ${\bf Z}^n$ is unique in its genus iff $n \leq 8$, and for $n = 9$ the genus consists of only ${\bf Z}^9$ and ${\bf Z} \oplus E_8$.
The comments indicate two ways to prove this, using $p$-neighbors (as implement in MAGMA) or the mass formula. Alternatively, one can use the fact that the theta function $\theta_L$ of any lattice $L$ in the genus of ${\bf Z}^n$ is a modular form of weight $n/2$ for an index-3 subgroup $\Gamma$ of the full modular group ${\rm PSL}_2({\bf Z})$.
For $n < 8$, there is only one choice of $\theta_L$ that has $q^0$ coefficient $1$, and we find that $L$ must have $2n$ vectors of norm $1$, and is thus isomorphic with ${\bf Z}^n$.
For $n=8$, either there are $16$ vectors of norm $1$ or there is a characteristic vector of norm zero. But in the latter case $L$ is an even lattice, and thus not in the genus of ${\bf Z}^8$. So we're back to having enough short vectors to identify $L$ with ${\bf Z}^8$.
The case of $n=9$ requires a bit more work. We can always write $L = {\bf Z}^m \oplus L_0$ where $L_0$ is a unimodular lattice of rank $n-m$ with no vectors of norm $1$. Using $\theta_{L_0}$ we soon find that $L_0$ is either $E_8$ or the trivial lattice of rank zero. Hence $L$ is either ${\bf Z} \oplus E_8$ or ${\bf Z}^9$.
For the details, see for example my papers
A characterization of the ${\bf Z}^n$ lattice, Math. Research Letters 2 (1995), 321-326 (arXiv: math.NT/9906019).
Lattices and codes with long shadows, Math. Research Letters 2 (1995), 643-651 (arXiv: math.NT/9906086).
Of course for every $n \geq 10$ the genus still contains at least the two lattices ${\bf Z}^n$ and ${\bf Z}^{n-8} \oplus E_8$, so $n=8$ is the last case where ${\bf Z}^n$ is unique in its genus.