According to this paper (which I've purchased though it is also visible as section 5 in this link), about 50% of the bromine is converted to bromate after 2 hours of exposure to sunlight with much of the rest converted to bromide. The conditions used were with seawater (which is around 67 mg/L Br2) and at a pH of 8 with around 4.5 ppm chlorine used in the experiment. Lower pH causes the decomposition and disproportionation reactions to occur more quickly, but without sunlight and at diluted concentrations in pools such reactions are slow. This link used de-ionized and raw water and found similar results with sunlight-induced bromate formation of 6.6% to 32% of bromide ion.
So if the amount of bromide added isn't too high, then keeping the pool chlorinated and exposing the water to sunlight should remove the bromide over time, probably in a few days if one uses typical sodium bromide algaecide treatments that are only a few ppm sodium bromide in dosage (but see below becaus your dose is much higher than that). Roughly take 1/4th the rate of chlorine demand and figure that's how much bromide is lost. In theory, the chlorine demand should be higher for a bromine pool since the bromine is unstabilized and the experiments seemed to show it decomposes fairly rapidly in sunlight.
The point about bromine tabs is relevant because the 5,5-dimethylhydantoin (DMH) in bromine tablets acts like CYA in binding to bromine so would slow down its degradation in sunlight.
3 pounds of sodium bromide in 10,000 gallons would be 56 ppm bromine. So this could take quite a while to remove.
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