You can figure out the amount of water in falling rain if you know the rate of accumulation (how much rain falls through a given volume in a given time, landing on the ground) and the velocity of the falling rain.

Here's a link to a post elsewhere on this same question. In that discussion, they assume a rainfall rate of 1 inch (2.54cm) per hour, and estimate a raindrop's terminal velocity (from observation) at about 9 m/s. They divide that rate by that velocity (converting units), and come up with about $\ce{3/4 cm^3}$ of water per $\ce{m^3}$ of rain.

So, given those assumptions, the density of (rain + air) is about .06% higher than the density of air without rain. Interestingly, that's much less than the difference in density between dry and humidity-saturated air (about 1% at 20°C). In other words, air at 100% relative humidity with heavy rain falling through it is still less dense than dry air.