Integral $\int_{-\infty}^\infty J^3_0(x) e^{i\omega x}\mathrm dx $
It turns out that the Fourier transform of $J_0^3$ can still be expressed in terms of complete elliptic integrals, but it's considerably more complicated than the formula for ${\cal FT}(J_0^2)$: for starters, it involves the periods of a curve $E$ defined over ${\bf C}$ but (except for a few special values of $\omega$) not over ${\bf R}$.
Assume $|\omega| < 3$, else $I(\omega) = 0$. Then the relevant curve is $$ E : Y^2 = X^3 - \bigl(\frac{3}{4} f^2 + \frac{27}{2} f - \frac{81}{4}\bigr) X^2 + 9 f^3 X $$ where $$ f = \frac12 \bigl( e + 1 + \sqrt{e^2-34e+1} \bigr) $$ and $$ e = \bigl( |\omega| + \sqrt{\omega^2-1} \, \bigr)^2. $$ Let $\lambda_1, \lambda_2$ be generators of the period lattice of $E$ with respect to the differential $dx/y$ (note that these are twice the periods that gp reports, because gp integrates $dx/2y$ for reasons coming from the arithmetic of elliptic curves). Then: if $|\omega| \leq 1$ then $$ I(\omega) = \left|\,f\,\right|^{5/2}\, \left|\,f-1\right| \frac{\Delta}{(2\pi)^2}, $$ where $\Delta = \bigl|{\rm Im} (\lambda_1 \overline{\lambda_2}) \bigr|$ is the area of the period lattice of $E$. If $1 \leq |\omega| \leq 3$ then $$ I(\omega) = \left|\,f\,\right|^{-4}\, \left|\,f-1\right|^5 (3/2)^{13/2} \frac{\Delta'}{(2\pi)^2}, $$ where $\Delta' = \bigl| {\rm Re}(\lambda_1 \overline{\lambda_2}) \bigr|$ for an appropriate choice of generators $\lambda_1,\lambda_2$ (these "appropriate" generators satisfy $|\lambda_1|^2 = \frac32 |\lambda_2|^2$, which determines them uniquely up to $\pm$ except for finitely many choices of $\omega$).
The proof, alas, is too long to reproduce here, but here's the basic idea. The Fourier transform of $J_0$ is $(1-\omega^2)^{-1/2}$ for $|\omega|<1$ and zero else. Hence the Fourier transforms of $J_0^2$ and $J_0^3$ are the convolution square and cube of $(1-\omega^2)^{-1/2}$. For $J_0^2$, this convolution square is supported on $|\omega| \leq 2$, and in this range equals $$ \int_{t=|\omega|-1}^1 \left( (1-t^2) (1-(|\omega|-t)^2) \right)^{-1/2} \, dt, $$ which is a period of an elliptic curve [namely the curve $u^2 = (1-t^2) (1-(|\omega|-t)^2)$], a.k.a. a complete eliptic integral. For $J_0^3$, we likewise get a two-dimensional integral, over a hexagon for $|\omega|<1$ and a triangle for $1 \leq |\omega| < 3$, that is a period of the K3 surface $$ u^2 = (1-s^2) (1-t^2) (1-(|\omega|-s-t)^2). $$ (The phase change at $|\omega|=1$ was already noted here in a now-deleted partial answer.) In general, periods of K3 surfaces are hard to compute, but this one turns out to have enough structure that we can convert the period into a period of the surface $E \times \overline E$ where $\overline E$ is the complex conjugate.
Now to be honest I have only the formulas for the "correspondence" between our K3 surface and $E \times \overline E$, which was hard enough to do, but didn't keep track of the elementary multiplying factor that I claim to be $\left|\,f\,\right|^{5/2}\, \left|\,f-1\right|$ or $\left|\,f\,\right|^{-4}\, \left|\,f-1\right|^5 (3/2)^{13/2}$. I obtained these factors by comparing numerical values for the few choices of $\omega$ for which I was able to compute $I(\omega)$ to high precision (basically rational numbers with an even numerator or denominator); for example $I(2/5)$ can be computed in gp in under a minute as
intnum(x=0,5*Pi,2*cos(2*x/5) * sumalt(n=0,besselj(0,x+5*n*Pi)^3))
There were enough such $c$, and the formulas are sufficiently simple, that they're virtually certain to be correct.
Here's gp code to get $e$, $f$, $E$, and generators $\lambda_1,\lambda_2$ of the period lattice:
e = (omega+sqrt(omega^2-1))^2
f = (sqrt(e^2-34*e+1)+(e+1)) / 2
E = ellinit( [0, -3/4*f^2-27/2*f+81/4, 0, 9*f^3, 0] )
L = 2*ellperiods(E)
lambda1 = L[1]
lambda2 = L[2]
NB the last line requires use of gp version 2.6.x; earlier versions did not directly implement periods of curves over $\bf C$.
For $\omega=0$ we have $e=1$, $f=3$, and $E$ is the curve $Y^2 = X^3 - 27 X^2 + 243 X = (X-9)^3 + 3^6$, so the periods can be expressed in terms of beta functions and we recover the case $\nu=0$ of Question 404222, How to prove $\int_0^\infty J_\nu(x)^3dx\stackrel?=\frac{\Gamma(1/6)\ \Gamma(1/6+\nu/2)}{2^{5/3}\ 3^{1/2}\ \pi^{3/2}\ \Gamma(5/6+\nu/2)}$? .
Actually more is true: the Fourier transform of $J_0(ax)J_0(x)^2$ for any $a\in \mathbb{R}$ can be expressed in elliptic integral $K(m) = \int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}$.
Here is the explicit form of the already nontrivial case $a=1$: let $$\tag{1}G(w) = \frac{2 \sqrt{4-3 t}}{\pi ^2 w(1-t)}K(m_+)K(m_-)$$ with $$2v = 5-w^2+\sqrt{(w^2-9)(w^2-1)} \qquad t=\frac{v}{v-1}$$ $$4m_{\pm} = 2\pm v\sqrt{4-v} -(2-v)\sqrt{1-v}$$ then for $s\in \mathbb{R}$, our Fourier transform is $$\color{royalblue}{\int_0^\infty J_0(t)^3 \cos(ts) dt} = -\lim_{\varepsilon\to 0^+}\Im[G(s+i\varepsilon)] $$
One has to be careful about branch problem though, see the end. Special cases: $$\begin{aligned}\int_0^\infty J_0(x)^3 \cos x dx &= \frac{(2-\sqrt{2}) (\sqrt{2}+1)}{16 \pi ^3}\Gamma \left(\frac{1}{8}\right)^2 \Gamma \left(\frac{3}{8}\right)^2 \\ \int_0^\infty J_0(x)^3 \cos(\sqrt{5}x)dx &= \frac{1}{4\pi^3} \Gamma \left(\frac{3}{20}\right) \Gamma \left(\frac{7}{20}\right) \Gamma \left(\frac{9}{20}\right) \Gamma \left(\frac{21}{20}\right) \end{aligned}$$
Forget our notation of $G$ above, let $$G(w) = \frac{1}{\pi^3}\int_{[0,\pi]^3} \frac{dx_1dx_2dx_3}{w-\cos x_1-\cos x_2 - \cos x_3}$$ then $G(w)$ is analytic in $\mathbb{C}-[-3,3]$. Using $\int_0^\pi e^{t\cos x} dx = \pi I_0(t)$ we have $$G(w) = \frac{1}{\pi^3}\int_{[0,\pi]^3} \int_0^\infty e^{-t(w-\cos x_1-\cos x_2 - \cos x_3)} dx_i dt = \int_0^\infty e^{-tw} I_0(t)^3 dt$$ here we assume $w > 3$ to ensure convergence. By deforming contour, we can switch the path of integral to be $\int_0^{i\infty}$ (integral over big circular arc $\to 0$ when $w>3$), so $$G(w) = i\int_0^\infty e^{-itw} J_0(t)^3 dt = -i\int_0^\infty e^{itw} J_0(t)^3 dt$$ Now RHS is analytic on $\Im w > 0$, so analytic continuation gives $$G(w) = -i\int_0^\infty e^{itw} J_0(t)^3 dt \qquad \Im w > 0$$ In particular, we see for $s\in \mathbb{R}$, $\lim_{\varepsilon\to 0^+}G(s+i\varepsilon)$ exists, denote it by $G_{\Re}(s) - iG_{\Im}(s)$. Then above equality gives $$G_{\Re}(s) = \int_0^\infty \sin(ts)J_0(t)^3 dt \qquad G_{\Im}(s) = \color{royalblue}{\int_0^\infty \cos(ts)J_0(t)^3 dt}$$ thus it suffices to show the triple integral actually equals RHS of $(1)$. The integral of $G_{\Re}(s)$ is not our goal, but we see it can be expressed as elliptic integral too.
The $G(w)$ is called Watson triple integral of simple cubic lattice. It turns out that the more general $$G(w,a) = \frac{1}{\pi^3}\int_{[0,\pi]^3} \frac{dx_1dx_2dx_3}{w-a\cos x_1-\cos x_2 - \cos x_3}$$ called Watson triple integral of singly anisotropic cubic lattice can also be expressed in terms of $K$. Its closed-form evaluation is a culmination of ingenious symbolic computation, which will be only briefly recapitulated below, the reader are strongly encouraged to read the reference for this spectacular achievement.
[to be added later....]
$(1)$ holds at least for $w>3$ when principal branch of square root and $K$ is taken. Since $G(w)$ has no branch point on $\Im w > 0$ from the triple integral definition, $(1)$ is true for all $w \in \mathbb{C}-[-3,3]$ after analytic continuation.
For our concerned range $0<\Re w < 3, \Im w >0$, I think the only term that can cause deviation from principal branch is the term $\sqrt{(w^2-9)(w^2-1)}$ in $v$. After this has been considered, the following Mathematica code can calculate $G(w)$ in our range ($\Im w = 0$ not included)
G[w_] := Module[{x, y, v, t, m1, m2}, x = Re[w]; y = Im[w];
If[5 x - x^3 + x*y^2 <= 0,
v = 1/2 (5 - w^2 + Sqrt[(w^2 - 9) (w^2 - 1)]),
v = 1/2 (5 - w^2 - Sqrt[(w^2 - 9) (w^2 - 1)])]; t = v/(v - 1);
m1 = 1/2 (1 + v*Sqrt[4 - v]/2 - (2 - v) Sqrt[1 - v]/2);
m2 = 1/2 (1 - v*Sqrt[4 - v]/2 - (2 - v) Sqrt[1 - v]/2);
2/Pi^2/w*Sqrt[4 - 3 t]/(1 - t)*EllipticK[m1] EllipticK[m2]];