There are two forces acting on the gripping edges of your G-clamp: the pressure difference between the inside water and the outside air, and the membrane stress of the bottle itself. Here is a sketch:
$\alpha$ is the angle between bottle and G-clamp edge: When the bottle does not bend at the G-clamp edge, $\alpha=90°$. The G-clamp edge is assumed to be circular with diameter $d$. (This can be easily generalized if required.)
The force on the G-clamp $F_{\mathrm{clamp}}$ can be calculated by
$$
F_{\mathrm{clamp}}=\frac{\pi}{4}d^2 \, \Delta p + \sin(\alpha) \, \sigma\, \pi\,d\,t
$$
with pressure difference between water and air $\Delta p$, membrane stress inside the bottle $\sigma$ and bottle wall thickness $t$.
The membrane stress $\sigma$ can be calculated using the formula for thin-walled pressure vessels:
$$
\sigma=\frac{\Delta p \, R}{2 \, t}
$$
with $R$ being the local bending radius at the clamp (or very close to it).
In total:
$$
F_{\mathrm{clamp}}= \left( 1 + 2 \, \sin(\alpha) \, \frac Rd \right) \frac{\pi}{4}d^2 \, \Delta p
$$
The difficult task is to determine $\alpha$ and $R$. However, if you can ensure $\alpha=90°$, i.e. the bottle does not bend at the G-clamp edge, then $\sin \alpha = 0$ and the clamp force is directly converted to pressure inside the bottle. This can be achieved by placing the bottle between two boards larger than the bottle.
Note: The hydrostatic pressure distribution is only important for very large bottles. The equation in your question ($\Delta p=\rho\,g\,z$ with $\Delta p$ being the pressure difference to $z=0$, vertical position $z$ pointing downward and fluid density $\rho$) is correct. For $\rho_{\mathrm{Water}}\approx 1000\,\mathrm{\frac{kg}{m^3}}$ and $g_{\mathrm{Earth}}\approx9,81\,\mathrm{\frac{N}{kg}}$, $\Delta p=1\,\mathrm{bar}=100\,000\,\mathrm{\frac{N}{m^2}}$ is reached for $z\approx 10.2\,\mathrm{m}$. If you feel it is important for your problem, you can superpose the hydrostatic pressure with the pressure calculated above.
