A sequence bounded in $L_p$, $p>1$, is uniformly integrable. Using this fact with $p=4/3$, it follows that $(f_n^3)$ is uniformly integrable. It now follows from Vitali's Convergence Theorem that $f_n$ converges to $f$ in the $L_3$-norm.


You can show that the sequence is uniformly integrable and arrive at your conclusion.


Using Fatou's Lemma, $$ \|f\|_{L^4}\le\liminf_{n\to\infty}\|f_n\|_{L^4}\tag{1} $$

Since $f_n$ converges pointwise, for any $\epsilon>0$ Egorov's Theorem states that except on $E_{\large\epsilon}$, a set of measure $\epsilon$, $f_n\to f$ uniformly. Therefore, $$ \lim_{n\to\infty}\int_{\large E_{\Large\epsilon}^c}\left|f_n(x)-f(x)\right|^3\,\mathrm{d}x=0\tag{2} $$ Furthermore, by Hölder's Inequality with $p=4/3$ and $q=4$, we have $$ \limsup_{n\to\infty}\int_{\large E_{\Large\epsilon}}\left|f_n(x)-f(x)\right|^3\,\mathrm{d}x\le\epsilon^{1/4}\sup_n\|f_n-f\|_{L^4}^3\tag{3} $$ Since $\epsilon$ is arbitrary, we get $$ \lim_{n\to\infty}\int_0^1\left|f_n(x)-f(x)\right|^3\,\mathrm{d}x=0\tag{4} $$