Here's a partial answer: For $p$ an even integer, it's true. Maybe someone else can see how to handle the other cases.
In this case, the desired integral can be written as $$\left(\sum_{k=0}^{p-1} (-1)^{k+1} \binom{p}{k} \int_{S^1} f_n^k f^{p-k} \phi\right) - \int_{S^1} f^p \phi.$$
Recall the following fact: for any $q>1$, if $g_n$ is bounded in $L^q$ and converges almost everywhere to $g$, then it converges to $g$ weakly in $L^q$. For the proof, see Bogachev's Measure Theory, or work it as a nice exercise.
Note that since $f_n$ converges weakly in $L^p$, by the uniform boundedness principle it is bounded in $L^p$ norm.
Now for any $1 \le k < p$, $f_n^k$ is bounded in $L^{p/k}$ norm. Therefore it converges, almost everywhere and weakly in $L^{p/k}$, to $f^k$. Moreover, $f^{p-k} \phi$ is in $L^{p/(p-k)}$, where $p/(p-k)$ is the conjugate exponent of $p/k$. So we have $\int_{S^1} f_n^k f^{p-k}\phi \to \int_{S^1} f^p \phi$, and the desired integral converges to $$ \left(\sum_{k=0}^{p-1} (-1)^{k+1} \binom{p}{k} - 1\right) \int_{S^1} f^p \phi$$ and the expression in parentheses is $(1-1)^p = 0$.