Largest $A\subset \mathbb{F}_2^n$ such that no two $a\neq b$ in $A$ add to an element of $A.$ If such a set $A$ of size $m$ exists, all its admissible pairwise sums must lie in its complement, thus
$$
\binom{m}{2} \leq 2^n-m,
$$
which gives $$m\leq 2^{(n+1)/2}\qquad (1)$$.
Edit: The upper bound is wrong. See the wonderful comments and examples below the question and in the answers. I can only accept one answer, unfortunately.
Since $a+b=c$ is the same as $a+b+c=0$ in characteristic 2, a randomly chosen set of size roughly $2^{n/3}$ will have a solution with some constant positive probability.
For simplicity let $n=2k$. Consider the nonzero elements in $\mathbb{F}_{2^k}$ as representing the vectors in $\mathbb{F}_2^k$
and take "vectors" of the form $(x,1/x)$ in $\mathbb{F}_2^n.$
Take $$(x,1/x)+(y,1/y)=(z,1/z)$$
equivalently $$x+y+z=0,$$ and $$1/x+1/y+1/z=0$$
as the simultaneous equations we must avoid.
By the Newton-Girard formulas we equivalently want to avoid
$$x+y+z=0,$$ and $$x^2+y^2+z^2=0$$
but in characteristic 2 the second equation is just the equare of the first.
This seems to say, if I pick a primitive element $\alpha$ in the subfield defined above and define
$v(i)=(\alpha^i,\alpha^{-i}+\beta)$ where $\beta$ is a nonzero element in $\mathbb{F}_{2^n} \setminus\mathbb{F}_{2^k}$, the vectors 
$$\{ v(i): 0\leq 2^k-2\}$$ form the type of set we want with $m=2^{n/2}-1.$
Question: Is there a better construction which approaches (1).
 A: The condition $a\ne b$ can in fact be safely dropped: if $2\cdot A:=\{a+b\colon a,b\in A, a\ne b\}$ is disjoint from $A$, then $0\notin A$ (unless $A\subseteq\{0\}$) and therefore also the larger set $2A:=\{a+b\colon a,b\in A\}$ is disjoint from $A$.
Since $2A\cap A=\varnothing$ is equivalent to $a+b+c\ne 0\ (a,b,c\in A)$, sets satisfying this condition are known as capsets. Every capset is contained in a maximal (by inclusion) capset; thus, it suffices to study maximal capsets.
As remarked in my comment above, the largest possible size of a capset in $\mathbb F_2^n$ is $2^{n-1}$. Davydov and Tombak gave a very non-trivial and delicate classification of maximal capsets of size larger than $2^{n-2}+1$; their result implies, in particular, that any capset of size larger than $5\cdot 2^{n-4}$ is contained in the complement of an index-$2$ subgroup.
A: The upper bound is wrong. The problem is that there is no need for the pairwise sums to be distinct. However it is true that there must be at least $m-1$ distinct sums, which gives $m \leq 2^{n-1}$.
This is sharp as we can take $A$ to consist of all elements with nonzero first entry, so all the sums of two elements of $A$ have zero first entry and so do not lie in $A$.
