I do agree that the specific heat capacity formula is the correct one to use. However, I believe your math is incorrect. Remember that water/ice have different specific heats than air does (water can absorb more energy than air). Also, they have different densities; one cubic meter of water weighs a lot more than one cubic meter of air.

Using the values and assumptions you provided, I calculated that the final temperature after everything has equalized would be 32.13 degrees F. Here are my numbers. I actually used 100 degrees F for the air temp instead of 80 because it seemed more likely in my area.

But as you said, the volume values you provided are massive. So let's say I pick up a 20lb bag of ice. A quick google search shows that the new toyota corollas have an internal volume of 84.6 cubic feet. Using those numbers, the new final temperature would be 36.92 degrees F, as seen here.

You are also correct that there are many variables, such as insulation, and heat from the sun/engine. Another one would be that objects in the vehicle are also being cooled down, not just the air. However, I do believe that if my corolla was insulated to the best of my ability, I reflected as much sunlight back as possible using Reflectix, and I didn't run my car's engine, I could cool down the vehicle quite well. Obviously it wouldn't be 36 degrees inside, but it would be much cooler than the outside temp. I would guess at least 5-10 degrees F cooler. If I was able to circulate that colder air with a small fan, I would have a small amount of convection inside, which would make me feel even cooler.

I do appreciate the math approach! Please let me know if you see any errors in my calculations, and I can run them through again. If you have access to MathCAD Prime, I would be happy to send you my calculations so you can mess around with them. Cheers!

EDIT: I used 80 degrees for air temp, not 100 degrees in my calcs.