Another proof can be achieved by using the isomorphism $L^2[-\pi,\pi]\simeq\ell^2(\mathbb{N})$, where you map $e^{ikt}\mapsto \delta_k$ (there's a factor $\sqrt{2\pi}$ around, but it affect convergence or not to zero).
Then, for any $a\in\ell^2(\mathbb{N})$, fix $\varepsilon>0$ and choose $M$ such that $\sum_{k>M}|a_k|^2<\varepsilon^2$. Then, using Hölder,
$$
|\langle f_n,a\rangle|=\frac1n\,\left|\sum_{k=1}^{n^2}a_k\right|
\leq\frac1n\,\sum_{k=1}^{n^2}|a_k|
=\frac1n\,\sum_{k=1}^{M}|a_k|+\frac1n\,\sum_{k=M+1}^{n^2}|a_k|
\leq\frac1n\,\sum_{k=1}^{M}|a_k|\\ +\frac1n\,\left(\sum_{k=M+1}^{n^2}|a_k|^2\right)^{1/2}(n^2-M)^{1/2} \\
\leq\frac1n\,\sum_{k=1}^{M}|a_k|+\varepsilon.$$
Letting $n\to\infty$ we get $\limsup_n|\langle f_n,a\rangle|<\varepsilon$, and as $\varepsilon$ was arbitrary, we get $\lim_n\langle f_n,a\rangle=0$.
For the sequence $\frac1n(f_1+\cdots+f_n)$ we can estimate:
$$
\|\frac1n(f_1+\cdots+f_n)\|_2^2=\left\|\frac1n\,\sum_{j=1}^n\sum_{k=1}^{j^2}\frac{\delta_k}j\right\|_2^2=\left\langle\frac1n\,\sum_{j=1}^n\sum_{k=1}^{j^2}\frac{\delta_k}j,\frac1n\,\sum_{j=1}^n\sum_{k=1}^{j^2}\frac{\delta_k}j\right\rangle\\
=\frac1{n^2}\sum_{j=1}^n\sum_{k=1}^{j^2}\sum_{l=1}^n\sum_{h=1}^{l^2}\frac{\langle\delta_k,\delta_h\rangle}{lj}
=\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{n}\frac{\min\{l^2,j^2\}}{lj}\\
=\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{j}\frac{\min\{l^2,j^2\}}{lj}
+\frac1{n^2}\sum_{j=1}^n\sum_{l=j+1}^{n}\frac{\min\{l^2,j^2\}}{lj}
=\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{j}\frac{l^2}{lj}
+\frac1{n^2}\sum_{j=1}^n\sum_{l=j+1}^{n}\frac{j^2}{lj}\\
=\frac1{n^2}\sum_{j=1}^n\sum_{l=1}^{j}\frac{l}{j}
+\frac1{n^2}\sum_{j=1}^n\sum_{l=j+1}^{n}\frac{j}{l}
=\frac1{n^2}\sum_{j=1}^n\frac{j(j+1)}{2j}
+\frac1{n^2}\sum_{l=1}^n\sum_{j=1}^{l-1}\frac{j}l\\
=\frac1{n^2}\sum_{j=1}^n\frac{j+1}{2}
+\frac1{n^2}\sum_{l=1}^n\frac{(l-1)l}{2l}
=\frac1{n^2}\sum_{j=1}^n\frac{j+1}{2}
+\frac1{n^2}\sum_{l=1}^n\frac{l-1}2\\
=\frac1{n^2}\sum_{k=1}^nk=\frac{n(n+1)}{2n^2}=\frac12+\frac1{2n}
$$