Since for $|c|<1$, and $|x|\leq c|$, $|f'_n(x)|=n|x^{n-1}|\leq n |c|^{n-1}$ and the sequence $\{n c^{n-1}\}$ is bounded, $\{f_n\}$ is equi-continuous at each point of $(-1,1)$.
We don't have equi-continuity at $1$, otherwise we would be able to find $\delta>0$ such that for each $n$, $|1-x^n|\leq \frac 12$ for $|1-x|\leq \delta$. THis gives $x^n\geq \frac 12$ for all $n$, which is not possible. A similar argument applies for $-1$.
We don't have equi-continuity at $x_0$ where $|x_0|>1$. Otherwise $|x^n-x_0^n|\leq 1$ for all $n$ and $|x-x_0|\leq\delta$ for some $\delta$, hence taking $x_n=x_0+1/n$ for a $n$ large enough, $\left|\left(x_0+\frac 1n\right)^n-x_0^n\right|\leq 1$ hence $|x_0|^n\left|\left(1+\frac 1{n|x_0|}\right)^n-1\right|\leq 1.$ As $\left|\left(1+\frac 1{n|x_0|}\right)^n-1\right|\to |e^{1/|x_0|}-1|\neq 0$, we get a contradiction.
Conclusion: $\{f_n\}$ is equi-continuous on the points of $(-1,1)$, and fail in all the others.