Firstly, for every real symmetric matrix $M$, the normalized eigenvectors $\{v_i\}_{i=1}^n$ of $M$ span the whole space (See here for proof). Without loss of generality, we assume such vectors are jointly orthogonal as well.
This means that one can rewrite each vector $x$ as
\begin{equation}
x = \sum_{i=1}^n c_i v_i.
\end{equation}
As a result, we have
\begin{equation}
Mx = \sum_{i=1}^n c_i \lambda_i v_i,
\end{equation}
where concludes
\begin{equation}
\|Mx\|_2 = \sqrt{\sum_{i=1}^n\lambda_i^2c_i^2},
\end{equation}
since $\|v_i\|_2=1$, and due to orthogonality of $v_i$s, where $\lambda_i$s are eigenvalues of $M$.
Further, since $x\in \{-1, 1\}^n$, we have $\sum_{i=1}^n c_i^2= n$. As a result, we have
\begin{equation}
\|Mx\|_2 \leq \sqrt{n}|\lambda_{\max}|,
\end{equation}
where $\lambda_{\max}$ is the eigenvalue with maximum absolute value.
Finally, Greshorin's circle theorem (see here), expresses that the eigenvalues of a matrix $M$ lies one circles with $M$s diagonal elements as centers and absolute sum of non-diagonal elements as radius's. Hence, for this matrix, we have $|\lambda_{\max}|< n$.
Finally, using above discussions we have
\begin{equation}
\|Mx\|_2 < n\sqrt{n}.
\end{equation}
My conjecture is that, the rate $O(n\sqrt{n})$ is optimal.
