Question:
For all $n$ large enough, there exists a matrix $A \in [-1,1]^{n \times n}$ such that $A\cdot A$ is a diagonal matrix with each diagonal entry at least $\frac{n}{100}$.
Discussion:
When there is a Hadamard matrix of order $n$, we can take $A$ to be that. But when $n$ is not a multiple of 4, there is no Hadamard matrix of order $n$.
Translation to Geometry: Let $B_n(r)$ be the open ball of radius $r$ centered at origin in $\mathbb{R}^n$. For all $n$ large enough, there are $n$ orthogonal vectors in $[-1,1]^n \setminus B_n\left(\sqrt{\frac{n}{100}}\right)$.
For example, you can use the Discrete Cosine matrix $$ C=[C_{ij}]_{1\leq i,j\leq n},\quad C_{ij}=\sqrt{\frac{2}{n}}\cos\left(\left(i-\frac{1}{2}\right) \left(j-\frac{1}{2}\right)\frac{\pi}{n} \right) $$ which is symmetric and self inverse, thus $$ C^\top C=I_n. $$ Take $A=\sqrt{n/2}\cdot C$ which should satisfy your requirement with each diagonal entry of $A^\top A$ at least $n/2$.