Without loss of generality, assume $x\ge0$ and $n\ge1$.
For $x\ge1$, the formula for the sum of a geometric series yields $ \begin{align} (1+x^{2n})^{\raise{2pt}{\large\frac{1}{2n}}}-x &=\frac{(1+x^{2n})-x^{2n}}{\sum\limits_{k=1}^{2n}(1+x^{2n})^{\raise{2pt}{\large\frac{k-1}{2n}x^{2n-k}}}}\\ &\le\frac{1}{2n}\tag{1} \end{align} $ For $x\le1$, the Mean Value Theorem says $ \begin{align} (1+x^{2n})^{\raise{2pt}{\large\frac{1}{2n}}}-1 &\le2^{\raise{2pt}{\large\frac{1}{2n}}}-1\\ &=e^\xi\left(\frac{1}{2n}\log(2)-0\right)\\ &\le\frac{\log(2)}{\sqrt{2}\,n}\tag{2} \end{align} $ for some $\xi\in(0,\frac{1}{2n}\log(2))$
Estimates $(1)$ and $(2)$ guarantee uniform convergence to $ f(x)=\left\{\begin{array}{} 1&\text{if }|x|\le1\\ |x|&\text{if }|x|>1 \end{array}\right. $