An $n×n$ matrix is formed using 0, 1 and -1 as its elements. The number of such matrices which are skew- symmetric, is

$3^{\frac{n (n-1)}{2}}$

All leading diagonal elements of a skew- symmetric are zero. So, to form a skew-symmketric matrix $A = [a_{ij}]$ of order $n×n$, we need to know the elements $a_{ij}$ for $i<j; i,j=1,2,...,n$. Each one of these $\frac{n^2 -n}{2}$ elements can take three values 0, 1, and -1. So, the number of skew-symmetric matrices is $3^{\frac{n^2 -n}{2}}=3^{\frac{n (n-1)}{2}}$.