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.
$\def\divides{\setminus}$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).