How many vectors can there be in $\mathbb{F}_2^{2n}$ such that no $n$ of them form a linearly dependent set? The bounds I have so far are embarrassingly far apart, though that probably means I should have thought about the question for longer before posting it.

To get an upper bound, observe that you can partition $\mathbb{F}_2^{2n}$ into $2^{n+2}$ translates of an $(n-2)$-dimensional subspace. If you choose more than $(n-1)2^{n+2}$ vectors, then $n$ of them must lie in one of those translates, and therefore in an $(n-1)$-dimensional subspace. So you definitely can't choose more than $Cn2^n$ vectors with the required property.

In the other direction, if you choose $M$ vectors randomly, then the probability that some fixed set of $n$ of them lives in an $(n-1)$-dimensional subspace is at most $n2^{-n}$ (since one of them must lie in the linear span of the others). So the expected number of problematic sets of size $n$ is at most $\binom Mn n2^{-n}$. If this is at most $M/2$, then we can get rid of a vector from each problematic set and we end up with no such sets. But for $n\binom Mn$ to be less than $2^n$ we basically need $M$ to be proportional to $n$, so this gives a lower bound of something like $2n$, which is pathetic as we could have just taken $2n$ linearly independent vectors.

I end up with a similarly pathetic bound if I try to pick vectors one by one, always avoiding the subspaces that the previous vectors require me to avoid.

I think I'm slightly more convinced by the lower bound, pathetic as it is. My rough reason is that the difficulty I run into feels pretty robust, and also that the result I prove in the upper bound is much stronger than it needs to be (since the subspace I obtain is essentially a translate of some fixed subspace). But basically I can't at the time of writing see even roughly what the bound should be.