$| f_n |^p - | f |^p - | f_n-f |^p$ converges in distribution sense if $f_k$ converges almost everywhere and weakly to $f$?

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 $-(1-1)^p = 0$.

Let $A_{n,\epsilon}$ be the set where $|f|<\epsilon|f_n|$. Now split the integral of $(|f_n|^p-|f|^p-|f_n-f|^p)\phi$ into an integral over $A_{n,\epsilon}$ and an integral over its complement. Since the $L^p$ norm of $f_n$ is bounded, the former part is bounded by a constant times $\epsilon$. The latter part converges to zero as $n\to\infty$ by the Lebesgue dominated convergence theorem.