Jeffrey - yes, such a calculation can be made, but keep in mind that the water pressure at that depth will have collapsed the tire, losing it's buoyancy. I'm not certain of the internal volume of the large low-pressure air tires used, but a 36 inch diameter torus with 12 inch width (roughly the same size I'm guessing) would displace 8527 cubic inches, or 37 gallons, or a bit more than 300# of seawater. That's 300 pounds of buoyancy, which may be enough for your scenario. It probably wouldn't float, that oleo strut and structure looks heavy, but it also wouldn't drop to the sea floor like an anchor. As it sank it would tend to bounce along with the current, becoming heavier and less mobile as water pressure collapsed the trapped air.
As I recall a rule of thumb, 2 feet of sea water depth produces about 1 psi increase in pressure. A 24psi tire would reach equilibrium at about 50 feet, and would begin to collapse (lose volume and buoyancy) below that depth.