Let $X_1, X_2,\dots$ be independent, uniform$(0,1)$ random variables.
By the law of large numbers we have
$$\begin{eqnarray*}
{X_1+\cdots + X_n\over n}&\to& \mathbb{E}(X)={1\over 2}\\
{X_1^2+\cdots + X^2_n\over n}&\to& \mathbb{E}(X^2)={1\over 3}\\
\end{eqnarray*}
$$
in probability as $n\to\infty$. Therefore
$${X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n}={X_1^2+\cdots +X^2_n\over n}\cdot{n\over X_1+\cdots +X_n}\to {2\over 3}$$ in probability as $n\to \infty$.
The ratio random variables ${X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n}$ are bounded below by zero and above by one. This guarantees convergence of the expectations, as well.
So $$\mathbb{E}\left({X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n}\right)\to{2\over 3}$$
which is the required result.
The probability approach makes things simpler. Let $X_1,X_2,\dots,X_n,\dots$ be independent random variables, which are uniform on $[0,1]$.
$X_1$ (or each other member in the family) has mean $\mu=\frac 12$ and variance $\sigma^2=\frac 1{12}$.
Let $Y_n$ be the mean $$Y_n=\frac 1n(X_1+X_2+\dots+X_n)\ .$$
Let $\mu$ be $1/2$ for short, and
$$
f(s) =
\frac 1{0!}f(\mu) +
\frac 1{1!}f'(\mu)(s-\mu) +
\frac 1{2!}f''(\mu)(s-\mu)^2 +
g(s)\ ,
$$
where $g(s)$ has an estimation of the shape $|g(s)|\le M\;|s-\mu|^3$ for some multiplicative constant $M$. We denote by $\Bbb E$ the expectation on the probability space where all variables $(X_n)$ live in.
Then let us denote by $I_n$ the integral:
$$
\begin{aligned}
I_n&=
\int_0^1\int_0^1\cdots\int_0^1 n
\left[\
f\left(\frac{x_1+\cdots+x_n}{n}\right)-f\left(\frac 12\right)
\ \right]
\,dx_1\,dx_2\cdots\,dx_n
\\
&=\Bbb E[\ n(f(Y_n)-f(\mu))\ ]
\\
&=\Bbb E\left[\ n\left(
\frac 1{1!}f'(\mu)(Y_n-\mu) +
\frac 1{2!}f''(\mu)(Y_n-\mu)^2 +
g(Y_n)
\right)\ \right]
\\
&=
\frac 1{1!}f'(\mu)\cdot n\;\underbrace{\Bbb E[\ (Y_n-\mu)\ ]}_{=0} +
\frac 1{2!}f''(\mu)\cdot n\;\underbrace{\Bbb E[\ (Y_n-\mu)^2\ ]}_{\operatorname{Var}[Y_n]=\sigma^2/n} +
\underbrace{n\;\Bbb E[\ g(Y_n)\ ]}_{\to 0}
\\
&\to\frac 1{2!}f''(\mu)\cdot\sigma^2=\frac 12f''(\mu)\cdot\frac 1{12}=\frac 1{24}f''(\mu)\ .
\end{aligned}
$$
Here, we have used the estimation for $g$, and the limit
$$
n\Bbb E[\ |Y_n-\mu|^3\ ]=
n\|\ (Y_n-\mu)^3\ \|_{L^1}
\le
n
\cdot
\|\ (Y_n-\mu)^1\ \|_{L^2}
\cdot
\|\ (Y_n-\mu)^2\ \|_{L^2}
\to 0\ .
$$
So the result is
$$
\lim_{n\to\infty}I_n=\frac 1{12}f''\left(\frac 12\right)\ .
$$
Best Answer
Using $$ \cos^2(x)=\frac{1+\cos(2x)}{2} $$ we get that $$ \begin{align} &\int_0^1\int_0^1\cdots\int_0^1\cos^2\left(\frac{a\pi}{2n}(x_1+x_2+\dots+x_n)\right)\,\mathrm{d}x_1\,\mathrm{d}x_2\dots\,\mathrm{d}x_n\\ &=\frac12+\frac12\mathrm{Re}\left(\int_0^1\int_0^1\cdots\int_0^1e^{\frac{ia\pi}{n}(x_1+x_2+\dots+x_n)}\,\mathrm{d}x_1\,\mathrm{d}x_2\dots\,\mathrm{d}x_n\right)\\ &=\frac12+\frac12\mathrm{Re}\left(\left[\int_0^1e^{\frac{ia\pi}{n}x}\,\mathrm{d}x\right]^n\right)\\ &=\frac12+\frac12\mathrm{Re}\left(\left[\frac{n}{ia\pi}\right]^n\left[e^{\frac{ia\pi}{n}}-1\right]^n\right)\\ &=\frac12+\frac12\mathrm{Re}\left(\left[\frac{2n}{a\pi}\sin\left(\frac{a\pi}{2n}\right)\right]^ne^{\frac{ia\pi}{2}}\right)\\ &=\frac12+\frac12\left[\frac{2n}{a\pi}\sin\left(\frac{a\pi}{2n}\right)\right]^n\color{#C00000}{\cos\left(\frac{a\pi}{2}\right)}\\ &\to\frac12+\frac12\cos\left(\frac{a\pi}{2}\right)\\ &=\cos^2\left(\frac{a\pi}{4}\right) \end{align} $$ If $a=1$, $\color{#C00000}{\cos\left(\frac{a\pi}{2}\right)}$ is $0$, so the integral is $\frac12$ for all $n$.