Let $q$ be a prime power and $\omega=\exp(2\pi i/q)$. For a fixed $y\in\mathbb{Z}_q^n$, the map
$$\mathbb{Z}_q^n\ni x\mapsto \omega^{x\cdot y}=\omega^{x_1y_1+\dots+x_ny_n}$$
is a character of $\mathbb{Z}_q^n$. The Hamming weight $\operatorname{wt}_H$ of $x\in\mathbb{Z}_q^n$ is the number of nonzero entries in $x$. Let $X_i=\{x\in \mathbb{Z}_q^n\mid \operatorname{wt}_H(x)=i\}.$ I want to prove that the sum $\sum_{x\in X_i}\omega^{x\cdot y}$ is constant for all $y\in X_k$. This is true for $q=2$ since the sets $X_i$ are invariant under the group operation of $S_n$ and for all $x,y\in X_k$ there exists $g\in S_n$ such that $gx=y$. Thus, we have
$$\sum_{x\in X_i}\omega^{x\cdot gy}=\sum_{x\in X_i}\omega^{g^{-1}x\cdot y}=\sum_{x\in X_i}\omega^{x\cdot y}.$$
But I don't see that the sum is constant if $q\geq 3$.
EDIT: I think that for $q\geq 3$ we could use a similar argument. If we take the group $S_{q-1}^n\wr S_n$, which operates on $x\in\mathbb{Z}_q^n$ by first permuting the positions of $x$ and then independently permuting the alphabet $\{1,2,\dots,q-1\}$ at each position, the sets $X_i$ should be invariant under this group action. Therefore, the sum should be constant for all $y\in X_k$. But I'm not completely sure that this is the right argument.