$\newcommand{\d}{\mathrm{d}}$ $\DeclareMathOperator{\Ai}{Ai}$
The equality
$$\int_0^{\infty}\Ai(z)^4\d z = \frac{\ln 3}{24\pi^2}$$
is established by Laurenzi (1993, Appendix A). To summarize, his strategy is to
- prove, using integration by parts and the Airy differential equation, that the integrals $$I_k\stackrel{\triangle}{=}\int_0^{\infty}z^k\Ai(z)^2\d z$$ satisfy the recurrence relation (Eq. 32) $$(2k+1)I_k= \tfrac{1}{2}k(k-1)(k-2)I_{k-3}$$ for all $k\geq 3$, whence $$I_k=\frac{(2/3)^{2/3}\Gamma(k+1)}{4\sqrt{3\pi}(12)^{k/3}\Gamma(k/3 + 7/6)};$$
- expand one of the factors of $\Ai(z)^2$ in the original integrand as a sum of hypergeometric $\,_2F_3$ series $$\Ai(z)^2=\frac{1}{12^{5/6}\pi^{3/2}}\sum_{k=0}^{\infty} (12z^3)^k\left(\frac{\Gamma(k+1/6)}{(3k)!}- 2(12)^{1/3}\frac{\Gamma(k+1/2)}{(3k+1)!}z+ 12^{2/3}\frac{\Gamma(k+5/6)}{(3k+2)!}z^2\right)$$ and integrate termwise to get $$\int_0^{\infty}\Ai(z)^4\d z = \frac{1}{72\pi^2}\sum_{k=0}^{\infty}\left(\frac{1}{k+1/6} - \frac{2}{k+1/2} + \frac{1}{k+5/6}\right)\text{;}$$ finally,
- express the infinite sum using the digamma function $\psi(z)$ and use that function's multiplication formula to reduce to the result.
Laurenzi, Bernard J. 1993. “Moment Integrals of Powers of Airy Functions.” Zeitschrift für Angewandte Mathematik und Physik 44 (5): 891–908. (ResearchGate 🔓)