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.   Note that for any $X\in \uso(n) $ we have

$$\dim \ker X equiv n\bmod 2. $$

If $k\equiv  n \bmod 2$ we have a diffeo $\uso(n)_k\to\uso(n-k)_0$ which associates to  a matrix $X\in\uso(n)_k$ its restriction to the orthogonal complement of the kernel.