As a slight extension of my comment, one can fix a (very) small subset $S\subset{\mathbb Z}_2^n$, and consider the addition Cayley graph, say $\Gamma$, induced by $S$ on ${\mathbb Z}_2^n$. (The vertices of $\Gamma$ are the elements of ${\mathbb Z}_2^n$, with two vertices adjacent whenever their sum is in $S$.) Now form an independent set $X\subset {\mathbb Z}_2^n$ by deleting from ${\mathbb Z}_2^n$, for every edge of $\Gamma$, one of the vertices incident with this edge. By the construction, we have $2X\subseteq{\mathbb Z}_2^n\setminus S$; furthermore, the expected density of $X$ is $2^{-|S|}$ and, typically, $X$ will be very uniformly distributed in ${\mathbb Z}_2^n$.

As a slight extension of my comment, one can fix a (very) small subset $S\subset{\mathbb Z}_2^n$, and consider the addition Cayley graph, say $\Gamma$, induced by $S$ on ${\mathbb Z}_2^n$. (The vertices of $\Gamma$ are the elements of ${\mathbb Z}_2^n$, with two vertices adjacent whenever their sum is in $S$.) Now form an independent set $X\subset {\mathbb Z}_2^n$ by choosing at random, for every edge of $\Gamma$, one of the two vertices incident with this edge, and deleting this vertex from ${\mathbb Z}_2^n$. By the construction, we have $2X\subseteq{\mathbb Z}_2^n\setminus S$; furthermore, the expected density of $X$ is $2^{-|S|}$ and, typically, $X$ will be very uniformly distributed in ${\mathbb Z}_2^n$.