Partial progress: It's easy to achieve $n-3$. Consider matrices of the form

$$\begin{pmatrix}
0 & 0 & r_1 & r_2 & \cdots & r_{n-3} & 0 \\
0 & 0 & 0 & r_1 & \cdots & r_{n-4} & r_{n-3} \\
r_1 & 0 & & & & & \\
r_2 & r_1 & & & & & \\
\vdots & \vdots & & & {\LARGE 0}& & \\
r_{n-3} & r_{n-4} & & & & & \\
0 & r_{n-3} & & & & & \\
\end{pmatrix}$$
where the bottom right square is entirely $0$. If this has rank $<4$, then the upper-left $4 \times 4$-submatrix implies $r_1^4=0$, so $r_1=0$. Then inductively $r_2^4=0$, and so forth.

For $n=5$, this only gives a $2$ dimensional subspace, and I argued in the comments on my other answer that a generic $3$ dimensional subspace of the $5 \times 5$ matrices should work. Right now, though, I can't see how to do better for $n \geq 5$.

Ah, slight improvement. For $n$ even, and taking advantage of the fact that I'm working over the reals, I can do $n-2$:
$$\begin{pmatrix}
0 & 0 & a_1 & b_1 & \cdots & a_{(n-2)/2} & b_{(n-2)/2} \\
0 & 0 & -b_1 & a_1 & \cdots & -b_{(n-2)/2} & a_{(n-2)/2} \\
a_1 & -b_1 & & & & & \\
b_1 & a_1 & & & & & \\
\vdots & \vdots & & & {\LARGE 0}& & \\
a_{(n-2)/2} & -b_{(n-2)/2} & & & & & \\
b_{(n-2)/2} & a_{(n-2)/2} & & & & & \\
\end{pmatrix}$$
If this has rank $<4$, then $(a_i^2+b_i^2)^2=0$ for all $i$, so over the reals this can only happen when it is $0$.

A potential strategy: What are (up to conjugation) the maximal subspaces of $n \times n$ skew-symmetric matrices on which the rank is always $\leq 4$? Then we can focus our efforts on finding large subspaces on each of these which miss the rank $2$ locus.

So far, I have only been able to find four maximal subspaces. I'll describe them all as block matrices with the size and nature of the blocks indicated:
$$\begin{pmatrix}
5 \times 5 & 0 \\
0 & 0
\end{pmatrix}$$
$$\begin{pmatrix}
0 & 2 \times (n-2) \\
(n-2) \times 2 & 0 \end{pmatrix}$$
$$\begin{pmatrix}
0 & 3 \times 3,\ \mbox{skew symmetric} & 0 \\
3 \times 3,\ \mbox{skew symmetric} & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}$$
$$\begin{pmatrix}
3 \times 3 & 0 & 0 \\
0 & 3 \times 3 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}$$

If these are the only options, then the only one which grows with $n$ is the second case so that's the one we need to concentrate on.

In that case, the question is equivalent to "what is the largest linear subspace of the $2 \times (n-2)$ matrices which includes no rank $1$ submatrices?" I can show that the answer to that question is $n-3$, over $\mathbb{C}$, and is $2 \lfloor (n-2)/2 \rfloor$ over $\mathbb{R}$; I'll post the argument if anyone cares.