Okay I made some rough estimates. Here's what I came up with.
First, forget about drift velocity. That should be replaced by velocity of the bob. Second, John Denker off-list mentioned I should also consider the 0.5CV^2 energy, but as best as I can tell that term is negligible, because C is so tiny (cf. numbers below).
Numbers (dropping factors of 2 or smaller):
magnetic field B = 100 gauss = 0.01 T
drop bob from an initial height L = 5 cm
bob is a cube of side length d = 1 cm
bob is made of Al which has resistivity p = 3e-8 ohm-m
permittivity of free space e0 = 9 pF/m
average speed of bob v = sqrt(g*L)
motional emf (or Hall voltage) V = B*v*d
resistance of bob R = p/d
capacitance of bob C = e0*d
dissipated energy E during one-quarter cycle = (V^2/R)*(L/v)
That works out to be B^2 g^0.5 L^1.5 d^3 / p = 1 uJ
compared to mechanical energy of mgL = 1 mJ (Al is 2.7 g/cm^3)
i.e. lose about 0.1% per quarter cycle. (And again, this continues only because the swinging velocity is ac not dc.)