It should be little surprise that this can be attacked via contour integration. As always, the challenge is to find the contour and the function to integrate over that contour. At the risk of presenting a deus ex machina, I am simply going to present these and demonstrate how they provide what we need.
Consider the following integral:
$$\oint_C dz \;f(z) $$
where
$$f(z) = \frac{e^{i \pi z^2}}{\sinh{\left (\sqrt{3} \pi z\right )} \left [ 2 \cosh{\left ( \frac{2 \pi}{\sqrt{3}} z\right )}-1\right ]}$$
and $C$ is the rectangle with vertices in the complex plane $\pm R\pm i \sqrt{3}/2$. This contour integral is thus equal to
$$\int_{-R}^R dx \; f\left (x-i \frac{\sqrt{3}}{2}\right ) + i \int_{-\sqrt{3}/2}^{\sqrt{3}/2} dy \, f(R+i y) \\ \int_R^{-R} dx\; f\left (x+i \frac{\sqrt{3}}{2}\right ) +i \int_{\sqrt{3}/2}^{-\sqrt{3}/2} dy \, f(-R+i y)$$
It should be clear that the second and fourth integrals vanish as $R\to\infty$, as these integrals vanish as $\pi e^{-\pi R}$ as $R\to\infty$. In this limit, then, the contour integral is equal to
$$\int_{-\infty}^{\infty} dx \; \left [ f\left (x-i \frac{\sqrt{3}}{2}\right ) - f\left (x+i \frac{\sqrt{3}}{2}\right )\right ] = i 2 e^{-i 3 \pi/4}\int_{-\infty}^{\infty} dx \frac{e^{i \pi x^2}}{2 \cosh{\left ( \frac{2 \pi}{\sqrt{3}} x\right )}+1}$$
I chose not to clutter up this space with the algebra involved in producing this last equation. The reader, however, should prove to his/herself that this is indeed correct.
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles of $f$ inside $C$. I leave it to the reader to verify that these poles are at $z=0$, $z=\pm i \sqrt{3}/3$, and $z=\pm i\sqrt{3}/6$. The residues at these poles are straightforward to compute because the poles are simple:
$$\operatorname*{Res}_{z=0} f(z) = \frac1{\sqrt{3} \pi}$$
$$\operatorname*{Res}_{z=\pm i \sqrt{3}/3} f(z) = \frac{e^{-i \pi/3}}{2 \sqrt{3} \pi}$$
$$\operatorname*{Res}_{z=\pm i \sqrt{3}/6} f(z) = -\frac{e^{-i \pi/12}}{2 \pi}$$
So the residue theorem states, equivalently, that
$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i \pi x^2}}{2 \cosh{\left ( \frac{2 \pi}{\sqrt{3}} x\right )}+1} &= \frac{e^{i 3 \pi/4}}{\sqrt{3}} + \frac{e^{i 5 \pi/12}}{\sqrt{3}} - e^{i 2 \pi/3}\\ &= \frac1{\sqrt{6}} (-1+i) + \frac1{2 \sqrt{6}} \left [ (\sqrt{3}-1) + i (\sqrt{3}+1)\right ] + \frac12 (1-i \sqrt{3})\end{align}$$
The integral of interest here is equal to $1/2$ the real part of the above, which is
$$\int_0^{\infty} dx \frac{\cos{\pi x^2}}{2 \cosh{\left ( \frac{2 \pi}{\sqrt{3}} x\right )}+1} = \frac{2+\sqrt{2}-\sqrt{6}}{8} $$
which was to be shown.
ADDENDUM
We may as well reap the reward of the imaginary part we get for free:
$$\int_0^{\infty} dx \frac{\sin{\pi x^2}}{2 \cosh{\left ( \frac{2 \pi}{\sqrt{3}} x\right )}+1} = \frac{\sqrt{2}+\sqrt{6}-2 \sqrt{3}}{8} $$
