We consider the $n$-dimenstional finite vector space $\mathbb{F}_2^{n}$ over the finite field of two elements. For a subset $A\subseteq \mathbb{F}_2^{n}$ of even size $|A|=2m$ and a linear form $l\in(\mathbb{F}_2^{n})^*$ let us say that $l$ bisects $A$, if $$|\{a\in A|l(a)=0\}|=|\{a\in A|l(a)=1\}|\,\,\,\,(=m). $$ Equivalently, for a subspace $L\leq \mathbb{F}_2^{n}$ of dimension $n-1$ (i.e. $|L|=2^{n-1}$), let us say that $L$ bisects $A$, if $$|A\cap L|=|A\setminus L|\,\,\,\,(=m). $$ How many subsets $A\subseteq \mathbb{F}_2^{n}$ of size $|A|=2m$ are there which are bisected by some suitable linear form (equivalently: by some subspace of dimension $n-1$)?

An alternative problem which I'm interested in (and which is maybe easier to solve??) is:

Determine the limit $$\lim_{n\rightarrow\infty}\frac{|\{A\subseteq \mathbb{F}_2^{n}|\,|A| \text{ is even and $A$ is bisected}\}|}{|\{A\subseteq \mathbb{F}_2^{n}|\,|A| \text{ is even}\}|}. $$ (Clearly, $|\{A\subseteq \mathbb{F}_2^{n}|\,|A| \text{ is even}\}|=2^{2^n-1}$.)

(My calculations seem to support the conjecture that the limit equals 1.)