How does L1 regularization make low-value features more zero than L2?
The penalty on coefficients of L1 is more aggressive on values close to zero than it is for L2.

With L1, when a weight value comes closer to zero, it tends to get even closer, because of the $\epsilon\lambda$ penalty which stays constant.

With L2, the $\epsilon\lambda w^{(t)}$ term gets smaller, so the regularization gets smaller as a weight comes closer to zero, with the update depending only on $\epsilon\Delta E(w)$.

So the main argument for this to my knowledge is that
$$
\lim_{w\rightarrow0}\lambda = \lambda\\
\lim_{w\rightarrow0}\lambda w = 0
$$