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
$$
\Im(G(a))=-\frac{\pi^3}{324}
$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{{4 \over 10} + {4 \cdot 7 \over 10 \cdot 20} +
{4 \cdot 7 \cdot 10 \over 10 \cdot 20 \cdot 30} + \cdots =
\sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{3k + 1} \over
\prod_{k = 1}^{n}10k}:\ {\LARGE ?}}$.
\begin{align}
&\bbox[10px,#ffd]{\ds{\sum_{n = 1}^{\infty}
{\prod_{k = 1}^{n}\pars{3k + 1} \over \prod_{k = 1}^{n}10k}}} =
\sum_{n = 1}^{\infty}
{3^{n}\prod_{k = 1}^{n}\pars{k + 1/3} \over 10^{n}n!} =
\sum_{n = 1}^{\infty}
{\pars{3/10}^{n} \over n!}\,\pars{4 \over 3}^{\large\overline{n}}
\\[5mm] = &\
\sum_{n = 1}^{\infty}
{\pars{3/10}^{n} \over n!}\,{\Gamma\pars{4/3 + n} \over \Gamma\pars{4/3}} =
\sum_{n = 1}^{\infty}
\pars{3 \over 10}^{n}\,{\pars{n + 1/3}! \over n!\pars{1/3}!}
\\[5mm] = &\
\sum_{n = 1}^{\infty}\pars{3 \over 10}^{n}\,{n + 1/3 \choose n} =
\sum_{n = 1}^{\infty}\pars{3 \over 10}^{n}
\bracks{{-4/3 \choose n}\pars{-1}^{n}}
\\[5mm] = &\
\sum_{n = 1}^{\infty}
{-4/3 \choose n}\pars{-\,{3 \over 10}}^{n} =
\bracks{1 + \pars{-\,{3 \over 10}}}^{-4/3} - 1
\\[5mm] = &\
\bbx{\pars{10 \over 7}^{4/3} - 1} \approx 0.6089
\end{align}
Best Answer
$\text{Lemma: } \sum_{k=0}^{n-1} k x^k=\frac{(n-1)x^n+1}{(x-1)}-\frac{(x^n-1)}{(x-1)^2} $$\text{Proof: }\sum_{k=0}^{n-1} x^k=\frac{x^n-1}{x-1}\implies\sum_{k=0}^{n-1} k x^{k-1}=\frac{n(x-1)x^{n-1}-(x^n-1)}{(x-1)^2}$
$$\sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot \frac{1}{\binom{2n}{r}}=\sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot(2n+1)\int_0^1t^{2n-r}(1-t)^{r}dt$$$$=-(2n+1)\int_0^1t^{2n}\sum_{r=1}^{2n-1}r\Big(1-\frac{1}{t}\Big)^rdt$$$$=-(2n+1)\int_0^1t^{2n}\bigg(\frac{(2n-1){\Big(1-\frac{1}{t}\Big)}^{2n}+1}{\frac{-1}{t}}-\frac{{\Big(1-\frac{1}{t}\Big)}^{2n}-1}{(\frac{-1}{t})^2}\bigg)dt$$
$$=(2n+1)\int_0^1\bigg({(2n-1)t{(1-t)}^{2n}+t^{2n+1}}+t^2{(1-t)}^{2n}-t^{2n+2}\bigg)dt$$
$$=(2n+1)\bigg(\frac{(2n-1)}{(2n+2)(2n+1)}+\frac{1}{2n+2}+\frac{2}{(2n+1)(2n+2)(2n+3)}-\frac{1}{2n+3}\bigg)=\frac{n}{n+1}$$
$\blacksquare$