No. If we take $\{-1/\sqrt n,1/\sqrt n\}^n$ instead of $\{0,1\}^n$, the problem reduces to asking if we can have many unit vectors $v_j$ with pairwise scalar products $-\gamma$ or less where $\gamma>0$ is a fixed number. But if we have $N$ such vectors, then the norm of the square of their sum is at most $N-\gamma N(N-1)$. Since this norm must be non-negative, we get $N-1\le\gamma^{-1}$ regardless of the dimension.

No. If we take $\{-1/\sqrt n,1/\sqrt n\}^n$ instead of $\{0,1\}^n$, the problem reduces to asking if we can have many unit vectors $v_j$ with pairwise scalar products $-\gamma$ or less where $\gamma>0$ is a fixed number. But if we have $N$ such vectors, then the square of the norm of their sum is at most $N-\gamma N(N-1)$. Since this square must be non-negative, we get $N-1\le\gamma^{-1}$ regardless of the dimension.