Here are two examples. I describe it as Lie algebras (over any field $K$).

(1) The 7-dimensional, 3-step nilpotent Lie algebra with basis $(X_1,\dots,X_7)$ and nonzero brackets

$$ [X_1,X_2]=X_4,[X_1,X_3]=X_5,[X_2,X_3]=X_6,[X_1,X_4]=[X_1,X_5]=[X_2,X_4]=[X_3,X_6]=X_7$$

(2) The 6-dimensional, 4-step nilpotent Lie algebra with basis $(X_1,\dots,X_6)$ and nonzero brackets

$$[X_1,X_2]=X_3, [X_1,X_3]=X_4, [X_2,X_3]=X_5, [X_1,X_5]=[X_2,X_4]=X_6.$$

Here's a common proof. In (1), let $V$ be the subspace generated by $X_1,X_2,X_3$, $W$ the subspace generated by $X_4,X_5,X_6$, and $Z$ the subspace generated by $X_7$. In (2), let $V$ be the subspace generated by $X_1,X_2$, $W$ the subspace generated by $X_4,X_5$, and $Z$ the subspace generated by $X_6$.

Then in each $\mathfrak{g}$ of these two: $Z$ is the 1-dimensional center, and the bracket $V\times W\to Z$ defines a non-degenerate pairing. Thus for each hyperplane $X$ in $V$, its centralizer in $W$ is equal to its orthogonal $X^\bot$ with respect to this pairing.

Each codimension 1 ideal of $\mathfrak{g}$ has the form $I_X=X\oplus \mathfrak{g}$ where $X$ is some hyperplane, and the center of $I_X$ is equal to the 2-dimensional ideal $X^\bot\oplus Z$.

Both Lie algebra being defined over $\mathbf{Q}$, we can consider the corresponding lattice, which therefore answers your question.