I'm going to assume you have a set up like the one in the figure below. The air pressure inside of the container will increase as the air is compressed by the water until the pressure inside the container balances the water pressure at the mouth of the container. Let $h$ be the depth of the mouth of the container below the water level. Then the water pressure at the mouth of the container is $P_w=\rho_w gh$, where $\rho_w\approx 997 kg/m^3$ is the density of water, and $g\approx 9.8 m/s^2$ is the gravitational acceleration at the surface of the earth. If we assume that the volume of the air doesn't change too much (which is a reasonable assumption unless you are very deep underwater), then you don't have to worry about the fact that the water pressure will lessen as water rises in the container. In this approximation, the pressure of the compressed air will be increased by $\rho_w gh$.
Just to plug in some numbers, if the mouth of the container is one meter underwater, then the pressure will increase by about $9770.6$ Pascals. For comparison, this is about a 10% increase from atmospheric pressure, which is 101,325 Pascals. The air pressure in the container will be equivalent to what you would feel if you dove one meter under water. The volume of the air in the container would decrease by about 10%, so the approximation that the height of the water doesn't change too much should hold as long as the container isn't too tall. A height of one meter would lead to about a 10% error in the calculated increase in pressure.