Another elementary proof:
Assume that $(x_n)$ is not weakly null. Then, passing to a subsequence, we get a functional $x^*$ such that $\inf_n |x^*(x_n)|>0$. By unconditionality, there are  $c_1$, $c_2$ $c_3>0$ such that 
$$
\Vert \sum_n a_n x_n\Vert \ge
c_1 \Vert \sum_n \epsilon_n a_n x_n\Vert \ge
c_2 \vert \sum_n  \epsilon_n a_n  x^*(x_n) \vert=
c_2  \sum_n  \vert a_n  x^*(x_n) \vert
 \ge c_3 \sum_n \vert a_n\vert,
$$
where $(\epsilon_n)$ is a suitable choice of signs.