show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$
Let $$f(y) = \int_{0}^{\infty} \frac{\sin^3{yx}}{x^3} \mathrm{d}x$$ Then, $$f'(y) = 3\int_{0}^{\infty} \frac{\sin^2{yx}\cos{yx}}{x^2} \mathrm{d}x = \frac{3}{4}\int_{0}^{\infty} \frac{\cos{yx} - \cos{3yx}}{x^2} \mathrm{d}x$$ $$f''(y) = \frac{3}{4}\int_{0}^{\infty} \frac{-\sin{yx} + 3\sin{3yx}}{x} \mathrm{d}x$$ Therefore, $$f''(y) = \frac{9}{4} \int_{0}^{\infty} \frac{\sin{3yx}}{x} \mathrm{d}x - \frac{3}{4} \int_{0}^{\infty} \frac{\sin{yx}}{x} \mathrm{d}x$$
Now, it is quite easy to prove that $$\int_{0}^{\infty} \frac{\sin{ax}}{x} \mathrm{d}x = \frac{\pi}{2}\mathop{\mathrm{signum}}{a}$$
Therefore, $$f''(y) = \frac{9\pi}{8} \mathop{\mathrm{signum}}{y} - \frac{3\pi}{8} \mathop{\mathrm{signum}}{y} = \frac{3\pi}{4}\mathop{\mathrm{signum}}{y}$$ Then, $$f'(y) = \frac{3\pi}{4} |y| + C$$ Note that, $f'(0) = 0$, therefore, $C = 0$. $$f(y) = \frac{3\pi}{8} y^2 \mathop{\mathrm{signum}}{y} + D$$ Again, $f(0) = 0$, therefore, $D = 0$.
Hence, $$f(1) = \int_{0}^{\infty} \frac{\sin^3{x}}{x^3} = \frac{3\pi}{8}$$
Use Parseval's theorem:
$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k)$$
where $f$, $g$ and $F$, $G$ are respective Fourier transform pairs, e.g.,
$$F(k) = \int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x}$$
etc. If $f(x) = \sin{x}/x$, then
$$F(k) = \begin{cases} \pi & |k| \le 1\\0 & |k| \gt 1 \end{cases}$$
Further, if $g(x) = \sin^2{x}/x^2$, then
$$G(k) = \begin{cases}\pi \left (1-\frac{|k|}{2} \right ) & |k| \le 2 \\ 0& |k| \gt 2\end{cases}$$
Then
$$\int_{-\infty}^{\infty} dx \,\frac{\sin^3{x}}{x^3} = \frac{1}{2 \pi} \int_{-1}^1 dk \, \pi^2 \left (1-\frac{|k|}{2} \right ) = \pi - \frac{\pi}{2} \int_0^1 dk \,k = \pi-\frac{\pi}{4}$$
Therefore
$$\int_{0}^{\infty} dx \,\frac{\sin^3{x}}{x^3} = \frac{3 \pi}{8}$$
ADDENDUM
You can also use contour integration techniques. For the integral
$$\int_0^{\infty} dt \frac{\sin^3{ \pi t}}{(\pi t)^3} \cos{u t}$$
I have derived a complete solution to the problem of its evaluation here using both contour integral techniques as well as the convolution theorem. You will see that the results agree for $u=0$ by a simple rescaling of the integral.
In this answer, the more general integral $$ \int_0^\infty\left(\frac{\sin(x)}{x}\right)^n\,\mathrm{d}x $$ is calculated.
Your integral is that integral for $n=3$.
A Different Way
In a fashion similar to this answer, we will use the equation $$ \frac{\mathrm{d}^2}{\mathrm{d}x^2}\frac{\sin^3(kx)}{k^3} =\frac{9\sin(3kx)-3\sin(kx)}{4k}\tag{1} $$ and the series for $0\lt x\le\pi$, $$ \sum_{k=1}^\infty\frac{\sin(kx)}{k}=\frac{\pi-x}{2}\tag{2} $$ Using $(2)$, we get $$ \begin{align} \sum_{k=1}^\infty\frac{9\sin(3kx)-3\sin(kx)}{4k} &=\frac94\frac{\pi-3x}{2}-\frac34\frac{\pi-x}{2}\\ &=\frac{3\pi}{4}-3x\tag{3} \end{align} $$ Integrating from $0$ twice to back out the derivatives taken in $(1)$ yields $$ \sum_{k=1}^\infty\frac{\sin^3(kx)}{k^3}=\frac{3\pi}{8}x^2-\frac12x^3\tag{4} $$ Set $x=1/n$ and multiply by $n^2$ to get $$ \sum_{k=1}^\infty\frac{\sin^3(k/n)}{k^3/n^3}\frac1n=\frac{3\pi}{8}-\frac1{2n}\tag{5} $$ and $(5)$ is a Riemann sum for $$ \int_0^\infty\frac{\sin^3(x)}{x^3}\,\mathrm{d}x=\frac{3\pi}{8}\tag{6} $$