Indeed, $\ell_1$, has a lot of extreme points. For example, take
$$
e^{(n)}=(\underbrace{0, \cdots, 0}_{n-1},1, 0, \cdots)
$$
And let $B$ be the closed unit ball in $\ell_1$. Clearly, $e^{(n)}\in B$. Now suppose that for $t \in (0,1)$, you have $b=\{ b_j \}_j , d=\{ d_j \}_j$ with $b,d \in B$ such that
$$
e^{(n)} = t b+ ( 1- t )d
$$
Then, you must have $1=tb_n+(1-t)d_n$ and $0=tb_j+(1-t)d_j$ for $j\neq n$ ,which gives that
$$
b_j=d_j=\left\{ \begin{array}{cc}
1 & \text{if } j=n \\
0 & \text{if } j\neq n \end{array} \right.
$$
Thus $b=d=e^{(n)}$, proving that $e^{(n)}$ is an extreme point of $B$ for all $n$.
Now you basically only need to answer if there are any more extreme points than the $
e^{(n)}$ to see all the extreme points of the space $\ell_1$.