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 $\frac{(n-1)n}{2} + n = \frac{n^2 + n}{2}$