Here's an elementary proof. Suppose $f(x) = x^k$ then we see that
\begin{align}
\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n})=&\ \int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ \left(\sqrt[n]{x_1\cdots x_n} \right)^k\\
=&\ \left(\int^1_0 dx\ x^{\frac{k}{n}} \right)^n = \left(1+\frac{k}{n} \right)^{-n}.
\end{align}
In particular, it follows
\begin{align}
\lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) = \lim_{n\rightarrow \infty}\left(1+\frac{k}{n} \right)^{-n} = \left(\frac{1}{e}\right)^k.
\end{align}
If $f$ is a polynomial, then it follows
\begin{align}
\lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) = f\left( \frac{1}{e}\right).
\end{align}
Next, if $f$ is continuous, then, by Wierestrass approximation theorem, there exists a sequence of polynomials $p_m$ such that $p_m \rightarrow f$ uniformly on $[0, 1]$. Finally, it follows
\begin{align}
\lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) =&\ \lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ \lim_{m\rightarrow \infty}p_m(\sqrt[n]{x_1\cdots x_n})\\
=&\ \lim_{n\rightarrow \infty}\lim_{m\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ p_m(\sqrt[n]{x_1\cdots x_n})\\
=&\ \lim_{m\rightarrow \infty}\lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ p_m(\sqrt[n]{x_1\cdots x_n})\\
=&\ \lim_{m\rightarrow \infty} p_m\left(\frac{1}{e} \right) = f\left(\frac{1}{e} \right).
\end{align}