Here is a small comment to Seva's answer, which allows to get smaller bounds for some fields.
Assume that we have two functions $f,g\in\mathbb F^A$ such that $$\sum_{a\in A} f(a)Q(a)=\sum_{a\in A} g(a)Q(a)=0 \tag{$\ast$} $$ for any polynomial $Q\in\mathbb F[x_1,\ldots,x_n]$ of degree $\deg Q\leqslant d$. Then $$ \sum_{a\in A,b\in A} f(a)g(b)Q(a,b)=0 $$ for any polynomial $Q$ of degree at most $2d+1$. (If $Q$ is a monomial, then our sum factorizes and one of factors is 0.)
Applying this with the polynomial $Q(x,y):=P(x-y)$, where $\deg P\leqslant 2d+1$ and $P(a,b)=0$ for $a,b\in A$, $a\ne b$, we get $$ 0=\sum_{a\in A,b\in A} f(a)g(b)P(a-b)=P(0)\cdot \sum_{a\in A} f(a)g(a). $$ Now if $P(0)\ne 0$, then $\sum_{a\in A} f(a)g(a)=0$ for any two functions $f,g\in\mathbb F^A$ . Then functions $f,g$ satisfying ($\ast$). This condition determines a linear subspace of $\mathbb F^A$ of dimension at least $|A|-\binom{n+d}d$ (or $|A|-\sum_{i\leqslant d} \binom{n}i$, if we consider only multilinear polynomials). We have thus shown that if $P(0,0)\ne 0$, then this is an isotropic subspace (any two vectors are mutually orthogonal.) The maximal dimension of an isotropic subspace is well-studied. It cannot exceed $|A|/2$ by obvious reasons (such a space is contained in its own orthogonal complement). For real field there is no isotropic subspace even of dimension $1$.
See the answer by Robin Chapman to my old question here for references on isotropic subspaces.
