If this were an even power it would be clear enough how to approach this, although I wouldn't be crazy about computing the Fourier transform.
For reference: I was able to solve $$ \int_{\mathbb{R}}\left( \frac{\sin x}{x}\right )^2 $$ using that $\frac{\sin x}{x}$ is the Fourier transform of the indicator function (interval dependent on one's definition of fourier transform).
Ideally a solution to the above would make reference to this fact assuming it is still germane.
Thanks!
Here are a few couple ways to get to the desired answer, using only standard techniques of Calculus
Solution 1
If you allow yourself to use it, this is a textbook example of when to use Integration by Parts
$$\int_{-\infty}^\infty \frac{\sin^3(x)}{x^3} dx = \,3\int_{-\infty}^\infty\frac{\sin^2(x)\cos(x)}{x^2}dx = \frac{3}{4}\int_0^\infty\frac{3\sin(3x)-\sin(x)}{x}dx$$
I assume you can take it from here by applying an integral you already know.
Solution 2
We can also apply Euler's Formula to convert our integral into
$$\Im\int_{-\infty}^\infty \frac{3e^{-ix}}{x^3}-\Im\int_{-\infty}^\infty \frac{3e^{ix}}{x^3}-\Im\int_{-\infty}^\infty \frac{e^{-3ix}}{x^3}+\Im\int_{-\infty}^\infty \frac{e^{3ix}}{x^3}$$
Each of which is trivial by using Differentiation Under the Integral Sign.
Note that we have to use the Principle Value here, as each of these doesn't normally converge
