I don't know if it's possible to determine the value of $\int_{0}^{\infty} \frac{\sin^{4}(x) }{x^{4}} \, dx$ from the value of $\int_{0}^{\infty} \frac{\sin^{4}(x) }{x^{2}} \, dx $.
But you mentioned that you used complex analysis to evaluate $\int_{0}^{\infty} \frac{\sin^{2}(x)}{x^{2}} \, dx$.
We can also use complex analysis to evaluate $\int_{0}^{\infty} \frac{\sin^{4}(x) }{x^{4}} \, dx$.
Using the trigonometric identity $ \displaystyle \sin^{4} x = \frac{1}{8} \Big(\cos 4x - 4 \cos 2x + 3 \Big)$, we get
$$ \begin{align} \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \, dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{4} x}{x^4} \, dx \\ &= \frac{1}{16} \int_{-\infty}^{\infty} \operatorname{Re} \, \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \ dx \\ &= \frac{1}{16}\int_{-\infty}^{\infty} \operatorname{Re} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, dx \\ &= \frac{1}{16} \, \operatorname{Re} \, \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, dx. \end{align}$$
So let's integrate the function $$f(z) = \frac{e^{4iz}-4e^{2iz}+3+4iz}{z^{4}}$$ around a contour the consists of the portion of the real axis from $-R$ to $R$, $R>0$, and the upper half of the circle $|z|=R$. To avoid the simple pole at the origin, the contour needs to be indented at the origin.
Letting the radius of the indentation go to zero and $R \to \infty$ (and applying Jordan's lemma along with the estimation lemma), we get
$$ \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, dx- i \pi \ \text{Res}[f(z),0] = 0,$$
where
$$ \operatorname{Res}[f(z),0] = \lim_{z \to 0} \frac{e^{4iz}-4e^{2iz}+3+4iz}{z^{3}} = \frac{16 \pi}{3} . $$
Therefore,
$$ \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \, dx = \frac{\pi}{3} .$$
Technically, it wasn't necessary to add $4ix$ to the numerator.
For reasons explained here, the Cauchy principal value of $ \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \, dx $ exists even though $\frac{e^{4iz}-4e^{2iz}+3}{z^{4}}$ has a pole of order $3$ at the origin.