First, if $e^N = \frac 12(x+y)$ with $x,y \in B_{\ell^1}$, then $1 = e^N_N = \frac 12(x_N + y_N)$ with $|x_N|, |y_N| \le 1$. Therefore $x_N = y_N = 1$ and since $\|x\|_1, \|y\|_1 \le 1$ we have $x=y= e^N$. So all $e^N$ are extreme points of $B_{\ell^1}$, so are $-e^N$.
Otherwise, if $x \in S_{\ell^1}$ with $x \ne \pm e^N$ for all $N$, there are $n \ne m$ with $x_n, x_m \ne 0$ and therefore 0 < |x_n|, |x_m| < 1, let $\eta := \min\{|x_n|, |x_m|, 1 - |x_n|, 1-|x_m|\}$ and set $y = x + \eta e^n - \eta e^m$ and $z = x - \eta e^n + \eta e^m$. Then $y,z \in B_{\ell^1}$, $x = \frac 12(y+z)$ and $y\ne z$. So $x$ is not extreme.
No let $x \in B_{\ell^1}$. We have \[ \sum_{i=1}^n x_i e^i = \sum_{i=1}^n |x_i|\cdot \mathrm{sgn} x_i\cdot e_i + \frac 12\left(1 - \sum_{i=1}^n |x_i|\right)e^{1} + \frac 12 \left(1 - \sum_{i=1}^n |x_i|\right)(-e^{1}) \] so $\sum_{i=1}^n x_i e^i \in\mathrm{conv}\mathrm{ext} B_{\ell^1}$ for each $n$. But $\sum_{i=1}^n x_ie^i \to x$ in norm, so we are done.