How to improve $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$

We have the elementary estimate $$1 \le \frac{z^4}{\sin^4 z} \le 1 + z^2 \varepsilon$$ where $$\varepsilon= \frac{\pi^2}{4} - \frac{4}{\pi^2}.$$ Let $z = (y/n)$ and multiply both sides by $\sin^4 y/y^4$. Then for $y \in [0,n \pi/2]$, one has: $$ \frac{\sin^4 y}{y^4} \le \left(\frac{\sin(y)/n}{\sin (y/n)}\right)^4 \le \frac{\sin^4 y}{y^4} + \frac{\sin^4 y}{y^2 n^2} \cdot \varepsilon $$ Make the substitution $x = y/n$ in the integral, it becomes

$$I_n:=n^2 \int^{n \pi/2}_{0} y \left(\frac{\sin(y)/n}{\sin(y/n)}\right)^4 dy$$

and thus

$$n^2 \int^{n \pi/2}_{0} \frac{\sin^4 y}{y^3} dy \le I_n \le n^2 \int^{n \pi/2}_{0} \frac{\sin^4 y}{y^3} dy + \varepsilon \cdot \int^{n \pi/2}_{0} \frac{\sin^4 y}{y} dy$$

The lower bound is asymptotic to $$n^2 \int^{\infty}_{0} \frac{\sin^4 x}{x^3} dx = n^2 \log 2,$$ and in fact since $$n^2 \int^{\infty}_{n \pi/2} \frac{\sin^4 x}{x^3} \le n^2 \int^{\infty}_{n \pi/2} \frac{1}{x^3} = \frac{2}{\pi^2} $$ one even has the lower bound $$I_n \ge n^2 \log 2 - \frac{2}{\pi^2}$$ On the other hand, an upper bound is given by $$ n^2 \int^{\infty}_{0} \frac{\sin^4 y}{y^3} dy = \varepsilon \cdot \int^{1}_{0} \frac{\sin^4 y}{y} + \varepsilon \cdot \int^{n \pi/2}_{1} \frac{1}{y} dy$$ $$ = n^2 \log 2 + \eta + \varepsilon \log(n \pi/2)$$

where $\eta \sim 0.160629\ldots$ and $\varepsilon \sim 2.062116\ldots$. From this you can obtain your explicit bound for $n \ge 3$ and check $n = 2$ by hand. Of course, it gives a more precise bound for larger $n$, and it's clear that you can push this much further if you want to.