Let's start with this Pfaff transformation for $\,a=\frac 14,b=\frac34,c=\frac23$ :
$$\tag{1}_2F_1\left(a,b\;;c\;;z\right)=(1-z)^{-a}\;_2F_1\left(a,c-b\;;c\;;\frac z{z-1}\right)$$
The 'Darboux evaluation' $(42)$ of Vidunas' "Transformations of algebraic Gauss hypergeometric functions" is :
$$\tag{2}_2F_1\left(\frac14,-\frac 1{12};\,\frac23;\,\frac {x(x+4)^3}{4(2x-1)^3}\right)=(1-2x)^{-1/4}$$
Solving $\,\displaystyle\frac {x(x+4)^3}{4(2x-1)^3}=\frac z{z-1}\,$ gives :
$$\tag{3}z=\frac{x(x+4)^3}{(x^2-10x-2)^2} $$
that we will use as :
$$\tag{4}z-1=\frac {4(2x-1)^3}{(x^2-10x-2)^2}$$
while $(1)$ and $(2)$ return :
$$_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)=\left[(z-1)(2x-1)\right]^{-1/4}$$
so that :
$$_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)=\left[\frac{4\,(2x-1)^4}{(x^2-10x-2)^2}\right]^{-1/4}$$
and (up to a minus sign) :
$$\tag{5}_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)^2=-\frac{x^2-10x-2}{2\,(2x-1)^2}$$
and indeed the substitution of $(z-1)$ and $_2F_1()^2$ with $(4)$ and $(5)$ in your formula gives :
$$27\,(z-1)^2\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^8+18\,(z-1)\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^4-8\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^2=1$$
Many other formulae of this kind may be deduced using Vidunas' paper.
A short proof.
$\displaystyle Q_0(x) :=\Gamma(x+1)=\lim_{n\to\infty}\frac{n^x}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)}~~ , ~~~~ Q_1(x) :=\lim_{n\to\infty}\frac{e^{xn}n^{-x^2/2}}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)^k}$
$\displaystyle \pi^{1/2} = Q_0\left(-\frac{1}{2}\right)~~ , ~~ A^3 = 2^{7/12}Q_1\left(-\frac{1}{2}\right)^2 ~~ , ~~ e^{G/\pi} = 2^{3/4}\left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right) }\right)^2$
It follows:
$\displaystyle \prod\limits_{k=1}^{2n}\left(1-\frac{1}{2k}\right)^{(-1)^k k} = \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{4k}\right)^{2k}}{\left(1-\frac{1}{4k-2}\right)^{2k-1}} = \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{2k}\right)^{2k-1} \left(1-\frac{1}{4k}\right)^{2k}}{ \left(1-\frac{3}{4k}\right)^{2k-1} } = $
$\displaystyle = \frac{ \prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right) n^{-1/2} }{ n^{-3/4} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right) } \left(\frac{ e^{-3n/4} n^{-9/32} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right)^k \prod\limits_{k=1}^n \left(1-\frac{1}{4k}\right)^k }{\prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right)^k e^{-n/2} n^{-1/8} e^{-n/4} n^{-1/32} }\right)^2$
$\displaystyle \to ~\frac{ Q_0\left(-\frac{1}{2}\right) }{ Q_0\left(-\frac{3}{4}\right) } \left(\frac{ Q_1\left(-\frac{3}{4}\right) }{ Q_1\left(-\frac{1}{2}\right) Q_1\left(-\frac{1}{4}\right) }\right)^2$
$\displaystyle = 2^{-1/6} \frac{ Q_0\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{1}{2}\right) } \cdot 2^{-7/12} Q_1\left(-\frac{1}{2}\right)^{-2} \cdot 2^{3/4} \left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right)}\right)^2$
$\displaystyle = \Gamma\left(\frac{1}{4}\right) 2^{-1/6} \pi^{-1/2} A^{-3} e^{G/\pi}$
Best Answer
Identities involving linear combinations of trigamma functions at rational arguments can be proved semi-automatically. The arguments here are of the form $\frac k{20}$ for $1\le k\le 19$, so denote $a_k=\psi_1(k/20)$ and write down some identities: $$a_k+a_{20-k}=\frac{\pi^2}{\sin^2k\pi/20},1\le k\le 10\tag{reflection}$$ $$a_5=\pi^2+8G\tag{special value}$$ $$a_k+a_{k+10}=4a_{2k},1\le k\le9\tag{duplication}$$ $$a_k+a_{k+5}+a_{k+10}+a_{k+15}=16a_{4k},1\le k\le4\tag{quadruplication}$$ $$a_k+a_{k+4}+a_{k+8}+a_{k+12}+a_{k+16}=25a_{5k},1\le k\le3\tag{quintuplication}$$ Treat the $a_k$ as variables and convert the identities into rows of a matrix equation $(a_1,a_2,\dots,a_{19},b)$, so e.g. reflection at $k=1$ becomes $$\left[\begin{array}{ccccccccccccccccccc|c}1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&\frac{\pi^2}{\sin^2\pi/20}\end{array}\right]$$ Now drop the last ($\mathbf b$) column and see if the desired linear combination $\mathbf c$ – in this case $(-1,0,1,6,0,0,0,10,0,0,0,0,0,0,0,0,0,0,0)$ – is in the row space of the remaining matrix $\mathbf A$, which can be done by trying to solve $\mathbf A^T\mathbf x=\mathbf c^T$. For the question's $\mathbf c$ there is a solution: $$\mathbf x=\left(\color{blue}{0, 0,\frac12, 0, 0, 2, 0, 2,\frac12, 0}, -12, \color{blue}{-\frac12, -2,\frac12, -2, 0, 0, 0, 0, 0}, 0, 0, 0, 0,\color{blue}{-\frac12, 0, 0}\right)^T$$ Then $\mathbf x\cdot\mathbf b$ gives an explicit expression for the trigamma linear combination: $$\frac12\frac{\pi^2}{\sin^23\pi/20}+2\frac{\pi^2}{\sin^26\pi/20}+2\frac{\pi^2}{\sin^28\pi/20}+\frac12\frac{\pi^2}{\sin^29\pi/20}-12(\pi^2+8G)$$ Simplifying shows this is equal to $-96G-\frac{24\pi^2\sqrt5}5+16\pi^2-2\pi^2\frac{15+\sqrt5}{\sqrt{10+2\sqrt5}}$ as suspected. (The simplification I get from Mathematica is $-96G+\pi^2(16-24/\sqrt5-2\sqrt{25-2\sqrt5})$).