I think the answer is no. To see this, observe that
\begin{equation*}
\left(\mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\right)^2=\mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\mathbb{E}_{\mathbf{y}}[(-1)^{u(\mathbf{y})}]=\mathbb{E}_{\mathbf{x},\mathbf{y}}[(-1)^{u(\mathbf{x})+u(\mathbf{y})}]=\mathbb{E}_{\mathbf{x},\mathbf{a}}[(-1)^{u(\mathbf{x})+u(\mathbf{x}+\mathbf{a})}],
\end{equation*}
where we use independence and then rewriting $\mathbf{y}=\mathbf{x}+\mathbf{a}$, and where all expectations are uniform over $\mathbb{F}_2^n$. The assumption then is that for $\mathbf{a}=\mathbf{0}$, the expectation over $\mathbf{x}$ is clearly $1$, while for all other values of $\mathbf{a}$, the expectation over $\mathbf{x}$ is $0$. It follows that
\begin{equation*}
\left(\mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\right)^2 = \frac{1}{2^n}\iff \mathbb{E}_{\mathbf{x}}[(-1)^{u(\mathbf{x})}] = \frac{\pm 1}{2^{n/2}}
\end{equation*}
However, it's not hard to see that this expectation must also be an integer multiple of $1/2^n$. But then this integer must be $\pm 2^{n/2}$, but this is integral if and only if $n$ is even.

additional conditionthat $n$ is odd. $\endgroup$