Denote by $\DeclareMathOperator{\uso}{\underline{so}}$ $\uso(n)$ the space of symmetric $n\times n$ matrices. The Cayley transform $\newcommand{\eC}{\mathscr{C}}$
$$ \uso(n) \ni X\mapsto \eC(X) := (1-X)(1+X)^{-1} $$
defines an diffeomorphism
$$ \eC:\uso(n)\to SO(n)^*:= \bigl\lbrace T\in SO(N);\;\; \ker(1+T)=0\;\bigr\rbrace. $$
Observe that $SO(n)^*$ is an open an dense subset of $SO(n)$.
Now observe that
$$ \forall X\in\uso(n):\;\;\ker X=\ker\bigl(\; 1-\eC(X)\;\bigr). $$
Thus the space $\uso(n)_k$ consisting of skew symmetric matrices $X\in\uso(n)$ such that $\dim \ker X= k$ has the same dimension as the space of orthogonal $n\times n$ matrices with the property that the eigenspace corresponding to the eigenvalue $1$ has dimension $k$. If $k\equiv n\bmod 2$ we deduce that this space the same dimension as the group $SO(n-k)$.
Hence
$$\dim \uso(n)_k= \binom{n-k}{2}. $$
The space you are interested is $\uso(n)_{n-4}$ and we deduce
$$\dim \uso(n)_{n-4}=\binom{4}{2}= 6. $$
Edit 1. I just realized that invoking the Cayley transform was not really needed.