Let
$\displaystyle \displaystyle I=\lim \int_0^1 |\sin(nx)|^3\text{ }dx$
Make the obvious substitution to get that
$\displaystyle \displaystyle I=\lim \frac{1}{n}\int_0^n |\sin(x)|^3\text{ }dx$
I can show it converges if that's what you really want, but one can show it converges, so we know that
$\displaystyle \displaystyle I=\lim \frac{1}{\pi n}\int_0^{\pi n}|\sin(x)|^3\text{ }dx$
But, of course we can rewrite this as
$\displaystyle \displaystyle I=\lim \frac{1}{\pi n}\sum_{j=0}^{n-1}\int_{j\pi}^{(j+1)\pi}|\sin(x)|^3\text{ }dx$
But, by the $\displaystyle \pi$-peridocity of $\displaystyle |\sin(x)|$ we know that
$\displaystyle \displaystyle \int_{j\pi}^{(j+1)\pi}|\sin(x)|^3\text{ }dx$
Is a constant function of $\displaystyle j$, and one can easily check by standard methods that
$\displaystyle \displaystyle \int_0^{\pi}|\sin(x)|^3\text{ }dx=\frac{4}{3}$
Thus,
$\displaystyle \displaystyle I=\lim \frac{1}{\pi n}\sum_{j=0}^{n-1}\int_{j\pi}^{(j+1)\pi}|\sin(x)|^3\text{ }dx=\lim \frac{4n}{3\pi n}=\frac{4}{3\pi}$
Remark: Of course, this method applied to a more general scenario shows that for sufficiently well-behaved functions ($\displaystyle C^0(\mathbb{R})$ should suffice) of period $\displaystyle T$ one has that
$\displaystyle \displaystyle \lim \int_a^b f(nx)\text{ }dx=\frac{b-a}{T}\int_0^T f(x)\text{ }dx$