**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: <a href="http://arxiv.org/abs/math/9906019">math.NT/9906019</a>).  
> Lattices and codes with long shadows, *Math. Research Letters* **2** (1995), 643-651 (arXiv: <a href="http://arxiv.org/abs/math/9906086">math.NT/9906086</a>).

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.