There have been many very good answers, so this is going to be another perspective of counting. We tackle the case of skew-symmetric first, which yields the condition that for any matrix $A$ with real entries, $a_{ij} = -a_{ij}$, so this means one side of the matrix is completely determined by the other as shown.

$$\begin{matrix}
\begin{pmatrix}
0 &*' \\
*& 0 \\
\end{pmatrix} & *
\end{matrix}$$

$$\begin{matrix}
\begin{pmatrix}
0 &*' & *' \\
*& 0 & *' \\
* & * & 0 \\
\end{pmatrix} & \begin{matrix}
*& \\
* & *
\end{matrix}
\end{matrix}$$

$$\begin{matrix}
\begin{pmatrix}
0 &*' & *' &*'\\
*& 0 & *' &*'\\
* & * & 0 &*'\\
* & * &* & 0
\end{pmatrix} & \begin{matrix}
* & & \\
*&* & \\
* & * &*
\end{matrix}
\end{matrix}$$

$$\begin{matrix}
\begin{pmatrix}
0&*'&*'&*'&\cdots \\
*&0&*'&*'& \\
*&*&0&*'& \\
*&*&*&0& \\
\vdots&&&&\ddots
\end{pmatrix} & \begin{matrix}
*& \\
*&*& \\
*&*&*& \\
\vdots&&&&\ddots
\end{matrix}
\end{matrix}$$

Notice that the triangle keeps getting larger and larger and at each stage and is incremented by $+1$. Since we have $n-1$ entries to consider we add up the number of entries, that is $$1+ 2 + \dots + (n-1).$$

So by Gauss, we have $$\frac{(n-1)(n-1+1)}{2} = \frac{(n-1)n}{2}$$ degree of freedom and that is the dimension we seek.

For the symmetric case we need to add back the $n$ objects in diagonal zeroed out, so $$\frac{(n-1)n}{2} + n = \frac{n^2 + n}{2}$$