You got me interested in some college history.

The formula for torque for a rod of circular cross-section is 1/2*pi*stress*R(cubed). Where stress is the strength of the material the rod is made of. I used 80,000 psi for the calcs, but it doesn't matter. R is the radius of the rod. For a hollow rod, it is simply 1/2*pi*stress*(R(cubed)-r(cubed)).
R is the outer radius, r is the inner radius.

A 7/8" diameter solid bar yields the same torque capability (876 foot-pounds, in this case) as a 1.25" bar with a .543 inner radius (1.086 inner diameter). Or stated another way, equal to a 1.25" hollow tube of wall thickness .082".

Interestingly, the hollow tube has a cross-section area of .3009 square inches. The solid bar has a cross-section area of .6013 square inches. Assuming the same length bars and ends (a bit of a stretch), the hollow bar would be almost 50% lighter.