Prove $_4F_3(1/8,3/8,5/8,7/8;1/4,1/2,3/4;1/2)=\frac{\sqrt{2-\sqrt2+\sqrt{2-\sqrt2}}+\sqrt{2+\sqrt2+\sqrt{2+\sqrt2}}}{2\,\sqrt2}$ How do I prove
$$_4F_3\left(\frac18,\frac38,\frac58,\frac78;\ \frac14,\frac12,\frac34;\ \frac12\right)\stackrel?=\frac{\sqrt{2-\sqrt2+\sqrt{2-\sqrt2\phantom{|}}}+\sqrt{2+\sqrt2+\sqrt{2+\sqrt2\phantom{|}}}}{2\,\sqrt2}$$

I encountered this conjecture when working with Fourier transforms of Kelvin functions.
 A: Some related values, based on Kirill's answer, given by radicals:
$$\begin{align}
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,\frac59\right) &= \frac{1}{4}\sqrt{9+\sqrt{30}+2\sqrt{15+3\sqrt{30}}}\\
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,\frac34\right) &= \frac{1}{2}\sqrt{4+\sqrt{3}+\sqrt{6+4\sqrt{3}}}\\
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,\frac89\right) &= \frac{1}{2}\sqrt{9+\sqrt{6}+\sqrt{12+6\sqrt{6}}}\\
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,9\right) &= \frac{1}{4}\sqrt{\frac{1}{2}-i\sqrt{2}+\sqrt{-6-6i\sqrt{2}}}
\end{align}
$$
It seems that in general for any $z$ rational values the solution is in the form of $\sqrt{x}$ where $x$ is a root of an $8$th-degree polynomial with integer coefficients.
A: $\def\divides{\setminus}$Interesting. First, write the hypergeometric function as the
following sum (with $z=\frac12$), using Gauss's multiplication theorem and
simplifying:
$$ F(z) = \sum_{n\geq0} \frac{\Gamma(\frac12+4n)z^n}{\sqrt\pi\Gamma(1+4n)}. $$
Define the function $f$ as the same thing but with $n$ instead of $4n$:
$$ f(z) = \sum_{n\geq0} \frac{\Gamma(\frac12+n)z^n}{\sqrt\pi\Gamma(1+n)} = \frac{1}{\sqrt{1-z}}, $$
and use the fact that when $\zeta=e^{2\pi i/4}=i$, 
$$\frac14\sum_{j=0}^3 \zeta^{jn} = [4\divides n], $$
to write
$$F(z) = \frac14\sum_{j=0}^3 f(z^{1/4}\zeta). $$
In other words, if $t_nz^n$ is the $n$-term of a sum, then
$$ \frac14\sum_{j=0}^3 t_n (z\zeta)^n = t_nz^n[4\divides n]. $$
The above gives a closed form for $F$ 
$$ F\big(z\big) = \frac14\left(\frac1{\sqrt{1-z^{1/4}}} + \frac1{\sqrt{1-i z^{1/4}}} + \frac1{\sqrt{1+iz^{1/4}}} + \frac1{\sqrt{1+z^{1/4}}} \right)\tag1 $$
Substituting $z=\frac12$, we find that
$$ F\big(\tfrac{1}{2}\big)=1.22198\dots$$
This (algebraic) number has the same 16th-degree minimal polynomial as the number that you gave, and they are numerically equal, so they are equal. (You can also do this by hand, it's not terribly hard).
