Finding $\int^{1}_{0}\frac{\ln^2(x)}{\sqrt{4-x^2}}dx$

A sketch (this approach is generalizable to arbitrary even powers of the logarithm):

We use the standardbranch of $\log$ throughout

$$ \Re \log^2(1-e^{2 ix})=\log^2(2 \sin(x))-(x-\frac{\pi}{2})^2 \quad (\star) $$

Integrating the LHS over a large rectangle in the upper half plane with verticies $\{(0,0),(\frac{\pi}{6},0),(\frac{\pi}{6},i\infty),(0,i\infty)\}$ we obtain (the integral over the imaginary line cancels, since we are only interested in real components)

$$ \Re\int_0^{\pi/6}\log^2(1-e^{2 ix})=-\Im\int_0^{\infty}\log^2(1-ae^{-2x})=-\Im(G(a)) $$

where $a=e^{-i \pi/3}$. Substituting $e^{-2x}=q$ we get $$ 2 G(a)=\int_0^1 \frac{\log^2(1-a q)}{q} $$

using repeated integration by parts we get ($\text{Li}_n(z)$ denotes the Polylogarithm of order $n$)

$$ 2 G(a)=-2\text{Li}_3(1-a)+2\text{Li}_2(1-a)\log(1-a)+\log(1-a)^2\log(a)+2\text{Li}_3(1) $$

adding some Polylogarithm wizardy we find that (see Appendix)

$$ \Im(G(a))=-\color{blue}{\frac{\pi^3}{324}} $$

using furthermore the trivial integral

$$ \int_0^{\pi/6}dx(x-\frac{\pi}{2})^2=\color{green}{\frac{19 \pi^3}{648}} $$

we find indeed

$$ \int_0^{\pi/6}dx\log^2(2 \sin(x))=\\ \color{green}{\frac{19 \pi^3}{648}}+\color{blue}{\frac{\pi^3}{324}}=\frac{7\pi^3}{216} $$

which is equivalent to the claim in question

Following OP we might rewrite the integral as an infinte sum, which gives us the hardly to believe corollary

$$ \sum_{n\geq 0}\frac{1}{16^n(2n+1)^3}\binom{2n}{n}=\frac{7\pi^3}{216} $$

Appendix

The functional equations of the Dilogarithm immediately delievers

$$\Re\text{Li}_2(1-a)= \frac{1}{2}(\text{Li}_2(1-a)+\text{Li}_2(1-a^{-1})\\=-\frac{1}{4}\log^2(a)=\frac{\pi^2}{36}$$

The Trilogarithm part is a bit trickier,

$\Im\text{Li}_3(1-z)=\Im\int_0^{1-z}\text{Li}_2(x)/x=-\Im\text{Li}_3(z)$ together with the inversion formula $\text{Li}_3(-z)-\text{Li}_3(-1/z)=-\frac{1}{6}(\log^3(z)+\pi^2\log(z))$gives us that

$$ \Im\text{Li}_3(1-a)=\frac{5\pi^3}{162} $$

since $\log(1-a)=\frac{i\pi}{3}$ we find

$$ 2\Im(G(a))=-2\frac{5\pi^3}{162}+2\frac{\pi^2}{36}\frac{\pi}{3}+\frac{\pi^2}9\frac{\pi}3 $$

or $$ \Im(G(a))=-\frac{\pi^3}{324} $$