First write \begin{align*} \log f_n(x) &= \sum_{j=1}^n \log(n^{x+1}+j^x) - \sum_{j=1}^n \log(n^{x+1}-j^x) \\ &= \sum_{j=1}^n \log\bigg( 1+\frac{j^x}{n^{x+1}} \bigg) + \sum_{j=1}^n \log\bigg( 1- \frac{j^x}{n^{x+1}} \bigg)^{-1}. \end{align*} The following inequalities are valid for all $0: \begin{align*} y-y^2 &< \log(1+y) < y \\ y &< \log(1-y)^{-1} < y+y^2 \end{align*} Therefore when $n\ge2$, \begin{align*} \log f_n(x) &> \sum_{j=1}^n \bigg( \frac{j^x}{n^{x+1}} - \bigg( \frac{j^x}{n^{x+1}} \bigg)^2 \bigg) + \sum_{j=1}^n \frac{j^x}{n^{x+1}} \\ &> 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x - \sum_{j=1}^n \frac1{n^2} \\ &= 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x - \frac1n \end{align*} and
\begin{align*} \log f_n(x) &< \sum_{j=1}^n \frac{j^x}{n^{x+1}} + \sum_{j=1}^n \bigg( \frac{j^x}{n^{x+1}} + \bigg( \frac{j^x}{n^{x+1}} \bigg)^2 \bigg) \\ &< 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x + \sum_{j=1}^n \frac1{n^2} \\ &= 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x + \frac1n. \end{align*} By the squeeze theorem, $ \lim_{n\to\infty} \log f_n(x) = 2 \lim_{n\to\infty} \sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x = 2 \int_0^1 u^x\, du = \frac2{x+1}, $ because the sum is a Riemann sum for the integral. In other words, $\lim_{n\to\infty} f_n(x) = e^{2/(x+1)}$ for all $x>0$ (since exponentiation is continuous).