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
$$\left(-\sum_{k=0}^{p-1} (-1)^{k} \binom{p}{k} \int_{S^1} f_n^k f^{p-k} \phi\right) - \int_{S^1} f^p \phi.$$

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*, Proposition 4.7.12, 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
$$ \left(-\sum_{k=0}^{p-1} (-1)^{k} \binom{p}{k} - 1\right) \int_{S^1} f^p \phi$$ 
and the expression in parentheses is $-(1-1)^p = 0$.