Now, I did the following, but it must be wrong since it doesn't depend on the choice of norm, so I get that $\displaystyle f_n \rightarrow 0$ in both cases.

Choose $\displaystyle x \in [0,\tfrac{1}{n}]$ for some small n.

$\displaystyle \exists \, n_0 > n$ st $\displaystyle x \notin [0,\tfrac{1}{n} ]$ but rather $\displaystyle x \in (\tfrac{1}{n} ,\tfrac{2}{n} )$

$\displaystyle \exists \, \tilde{n}_0 > n_0 $ st $\displaystyle x \notin (\tfrac{1}{n},\tfrac{2}{n} )$ but rather $\displaystyle x \in [\tfrac{2}{n} ,1 )$

$\displaystyle \Rightarrow \, \exists \, \hat{n}_0$ st $\displaystyle f_n = 0 \, \forall \, n \ge \hat{n}_0$

So,

clearly, we have that $\displaystyle \forall \, \epsilon >0 \, \exists \, \hat{n}_0 \in \mathbb{N}$ st $\displaystyle \parallel 0 - f_n \parallel _1 < \epsilon \, \forall \, n \ge \hat{n}_0$