Consider the addition Cayley graph $\Gamma$ induced by $F$ on $\mathbb Z_2^n$ (which is the graph with the vertex set $\mathbb Z_2^n$, with the vertices $u,v\in\mathbb Z_2^n$ adjacent whenever $u+v\in F$). A set $A$ with $(A+A)\cap F=\emptyset$ is an independent set in $\Gamma$. Since $\Gamma$ is regular of degree $|F|$, it has an independent set of size at least $2^n/(|F|+1)$.

In general, this bound is best possible: it is attained when $F$ is a subgroup of $\mathbb Z_2^n$ with the zero element removed (so that $|F|=2^k-1$, where $k$ is the rank of $F$), and $A$ contains a unique element from each $F$-coset.

Better bounds can be given if some information about the set $F$ is available. Say, if $F=\{f_1,\dotsc, f_m\}$ is an independent set, then one can find a subgroup $H<\mathbb Z_2^n$ such that $\mathbb Z_2^n=\langle F\rangle\oplus H$, and take
$$ A := \{ c_1f_1+\dotsb+c_mf_m+h\colon c_1+\dotsb+c_m=0,\ h\in H \} $$
to have $|A|=2^{n-1}$.

What is a non trivial lower bound...Are you sure you did not mean "upperbound"? $\endgroup$ – fedja May 2 at 22:57