What is the limit?

By tancelocation -

$\begin{aligned}\left(\int_0^1 \frac{1}{1+x^n} dx\right)^n & = & \left(1- \underbrace{\int_0^1 \frac{x^n}{1+x^n} dx}_{=:J_n}\right)^n \end{aligned}$

Using integration by parts you get $J_n = \frac 1n\int_0^1 x\frac{nx^{n-1}}{1+x^n} dx = \frac 1n\left(\ln 2 - \underbrace{\int_0^1 \ln (1+x^n) dx}_{=:I_n}\right)$

Now, DCT (or just squeezing) gives immediately $\lim_{n\to \infty }I_n = 0$

All together

$\begin{aligned}\left(\int_0^1 \frac{1}{1+x^n} dx\right)^n & = & \left(\left(1-\frac{\ln 2 - I_n}n\right)^{\frac{n}{\ln 2 - I_n}}\right)^{\ln 2 - I_n} \\ & \stackrel{n\to\infty}{\longrightarrow} & \left(e^{-1}\right)^{\ln2} = \frac 12 \end{aligned}$