This formula can actually be proved using only properties of the Gamma
function already known to Gauss, with no need to invoke special values of
Dirichlet series.  The relevant identities are
$$
\Gamma(z) \, \Gamma(1-z) = \frac\pi{\sin(\pi z)},
$$
already cited by **john mangual** as the "mirror formula", and the
*triplication formula* for the Gamma function, i.e. the case
$k=3$ of <a href="https://en.wikipedia.org/wiki/Multiplication_theorem#Gamma_function.E2.80.93Legendre_function">Gauss's
multiplication formula</a>:
$$
\Gamma(z) \, \Gamma\bigl(z+\frac13\bigr) \, \Gamma\bigl(z+\frac23\bigr)
= 2\pi \cdot 3^{\frac12-3z} \Gamma(3z)
$$
[the $k=2$ case is the more familiar duplication formula
$\Gamma(z) \, \Gamma(z+\frac12) = 2^{1-2z} \sqrt{\pi}\, \Gamma(2z)$].

Take $z=1/24$ and $z=1/8$ in the triplication formula, multiply,
and remove the common factors $\Gamma(1/8) \, \Gamma(3/8)$ to deduce
$$
\Gamma\bigl(\frac{1}{24}\bigr)
\Gamma\bigl(\frac{11}{24}\bigr)
\Gamma\bigl(\frac{17}{24}\bigr)
\Gamma\bigl(\frac{19}{24}\bigr)
= 4 \pi^2 \sqrt{3}.
$$
Take $z=5/24$ and $z=7/24$ in the mirror formula and multiply to deduce
$$
\Gamma\bigl(\frac{5}{24}\bigr)
\Gamma\bigl(\frac{7}{24}\bigr)
\Gamma\bigl(\frac{17}{24}\bigr)
\Gamma\bigl(\frac{19}{24}\bigr)
= \frac{\pi^2}{ \sin (5\pi/24) \sin (7\pi/24) }.
$$
Hence
$$
\frac{\Gamma(1/24) \, \Gamma(11/24)} {\Gamma(5/24) \, \Gamma(7/24)}
= 4 \sqrt{3} \sin (5\pi/24) \sin (7\pi/24),
$$
which is soon reduced to the radical form $\sqrt3 \cdot \sqrt{2+\sqrt3}$.