The volume of the cone is approximately $1/n$ for large $n$. Consider $C=N_S(e_n)$. In order for a vector $y$ to lie in $C$, it suffices that $y\cdot (-e_n)\le 0$ and $y\cdot (e_i-e_n)\le 0$ for each $i<n$.

That is, we require $y_n\ge 0$ and $y_n\ge y_i$ for each $i<n$. In other words, we need $y_n=\max(y_1,\ldots,y_{n-1})$ and $y_n\ge 0$.

The first condition occupies a proportion $1/n$ of the space (by symmetry) and given that the first condition holds, the second condition holds with probability close to 1.

The volume of the cone is approximately $1/n$ for large $n$. Consider $C=N_S(e_n)$. In order for a vector $y$ to lie in $C$, it suffices that $y\cdot (-e_n)\le 0$ and $y\cdot (e_i-e_n)\le 0$ for each $i<n$.

That is, we require $y_n\ge 0$ and $y_n\ge y_i$ for each $i<n$. In other words, we need $y_n=\max(y_1,\ldots,y_{n-1},y_n)$ and $y_n\ge 0$.

The first condition occupies a proportion $1/n$ of the space (by symmetry) and given that the first condition holds, the second condition holds with probability close to 1.