A modest start, but before doing so, one suggestion: I would index the columns and rows from $0$ to $n-1$, so $a_{ij} = (-1)^{\lfloor 2ij/n\rfloor}$.
Let $n=2m+1$ be an odd prime. Then $\lfloor 2kj/n\rfloor$ and $\lfloor 2(n-k)j/n\rfloor$ have different parity for $1\le j, k\le n-1$.
Let $v_k$ be the row vector with $1$ in positions $k$ and $n-k$ and $0$ elsewhere. Thus $v_k$ is a left eigenvector for $1\le k\le n-1$ of $A_n'$ where $A_n'$ is the matrix $A_n$ with the first column removed. As the first column of $A_n$ is constantly $1$, the vectors $v_k-v_m$ for $1\le k\le m-1$ are left eigenvectors of $A_n$. And clearly they are linearly independent.
Thus for $n$ an odd prime, we get $\operatorname{rank}(A_n)\le\lfloor(n+3)/2\rfloor$.