Some algebra to arrive at a manageable integral
Notice that the $\arccos $ argument is the "$+$" solution of a quadratic equation:
$$\cos x\left(2\sin^2x\pm\sqrt{1+4\sin^4x}\right)=\frac{2\sin x \sin(2x)\pm\sqrt{4\sin^2 x\sin^2(2x)+4\cos^2 x}}{2}$$
More exactly, with $b=-2\sin x\sin(2x)$ and $c=-\cos^2x$ in $y_{\pm}=\frac{-b\pm\sqrt{b^2-4c}}{2}$.
We also have $\arccos y_-\pm\arccos y_+=\arccos\left(y_-y_+\mp\sqrt{(1+y_-y_+)^2-(y_-+y_+)^2}\, \right)$, and since Vieta's relations allows us to utilize $y_-+y_+=2\sin x\sin(2x)$ and $\ y_-y_+=-\cos^2 x$, it makes sense to consider the following integrals:
$$I_{\pm}=\int_{\arccos(1/4)}^{\pi/2}\arccos\left(\cos x\left(2\sin^2x\pm\sqrt{1+4\sin^4x}\right)\right) dx$$
The original integral can then be rewritten as $I_+=\frac12\left((I_- + I_+)-(I_- - I_+)\right)$, where:
$$I_- \pm I_+=\int_{\arccos(1/4)}^{\pi/2}\arccos\left(-\cos^2 x\mp\sin^2 x\sqrt{1-16\cos^2 x}\right)dx$$
$$\overset{\sqrt{1-16\cos^2 x}\to x}=\int_0^1 \frac{x\arccos \left(-\frac{1-x^2}{16}\mp x\frac{15+x^2}{16}\right)}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx$$
Moreover, substituting $x\to -x$ in $I_- - I_+$, gives us:
$$I_+=\frac12\int_{-1}^1 \frac{x\arccos \left(-\frac{1-x^2}{16}- x\frac{15+x^2}{16}\right)}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx=\int_{-1}^1 \frac{x\operatorname{arctan}\sqrt{\frac{1-x^2}{15+x^2} \left(1+\frac{16}{(1-x)^2}\right)}}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx$$
In the last step it was utilized that $\arccos x =2\arctan \sqrt{\frac{1-x}{1+x}}$.
Evaluation of the integral
$$\int_{-1}^1 \frac{x\operatorname{arctan}\sqrt{\frac{1-x^2}{15+x^2} \color{red}{\left(1+\frac{16}{(1-x)^2}\right)}}}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx=\int_{-1}^1 \int_0^{\sqrt{\color{red}{f(x)}}} \frac{x}{(15+x^2)+(1-x^2) y^2}dy dx$$
$$\overset{(*)}=\int_{-1}^1 \int_0^{x} \frac{x\left(\sqrt{\color{red}{f(y)}}\right)'}{(15+x^2)+(1-x^2) \color{red}{f(y)}}dy dx =\int_{-1}^1 \int_0^x (*) dy dx = \int_{-1}^1 \int_y^1 (*) dx dy$$
$$=\frac12\int_{-1}^1 \frac{\ln\left(\frac{2}{1-y}\right)}{\sqrt{16+(1-y)^2}}dy\overset{\frac{1-y}{2}\to y}=-\frac12 \int_0^1 \frac{\ln y}{\sqrt{4+y^2}}dy \overset{y \to \frac{1-y^2}{y}}=\frac12 \int_{\large \frac{1}{\phi}}^1 \frac{\ln\left(\frac{y}{1-y^2}\right)}{y}dy $$
$$=\frac14\operatorname{Li}_2(1) - \frac14\operatorname{Li}_2\left(\frac{1}{\phi^2}\right) - \frac{1}{4}\ln^2\phi \overset{(**)}=\, \boxed{\frac{\pi^2}{40}}$$
In $(**)$, the following dilogarithm values were employed:
$$\operatorname{Li}_2(1)=\frac{\pi^2}{6},\quad \operatorname{Li}_2\left(\frac{1}{\phi^2}\right) = \frac{\pi^2}{15}-\ln^2\phi,\quad \phi=\frac{1+\sqrt 5}{2}$$
Also, the $(*)$ step is not necessary, however I didn't know how to change the order of integration directly, so I had to make the substitution $y^2\to f(y)$ after noticing that the $x$-integral is odd which allows to write $\int_{-1}^1 \int_0^{f(x)}dydx = -\int_{-1}^1 \int_{f(x)}^\infty dydx$. Finally, $-\int_{-1}^1 \int_x^1 dydx$ was also rewritten as $\int_{-1}^1 \int_0^x dydx$ in order to change the order of integration easier.
Generalization
By the same approach, for $a\in(0,1)$, we can also obtain:
$$\int_{\arccos \frac{a}{2}}^\frac{\pi}{2} \arccos\left(\cos x\left(\frac{\sin^2 x}{a}+\sqrt{1+\frac{\sin^4 x}{a^2}}\right)\right)dx$$
$$=\boxed{\frac{\pi^2}{24}-\frac{\operatorname{arcsinh}^2(a)}{4}+\frac{\ln(2a)\operatorname{arcsinh}(a)}{2}-\frac{1}{4}\operatorname{Li}_2\left(\frac{1}{(a+\sqrt{1+a^2})^2}\right)}$$
