Prove $_4F_3(1/8,3/8,5/8,7/8;1/4,1/2,3/4;1/2)=\frac{\sqrt{2-\sqrt2+\sqrt{2-\sqrt2}}+\sqrt{2+\sqrt2+\sqrt{2+\sqrt2}}}{2\,\sqrt2}$

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

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.