It's probably because the Ksp of Ba(NO3)2 >>> BaSO4, Ksp of Ba(NO3)2/Na2SO4/NaNO3 > 1 (all 3 compounds are totally soluble in water), and \[Na2SO4\] >>> \[Ba(NO3)2\]. If that's the case, then via chatelier's principle, the reaction will go very forward into completion because all the \[SO4\] is available to react with the \[Ba\] and the Ksp becomes dependent only on \[Ba\]. All the other species don't affect the Ksp because they're all soluble. In other words, Ksp = \[Ba\]\^2 when \[SO4\] no longer becomes a limiting reagent.

I see your point! But let’s imagine that the Ksp of BaSO4 is 10^-10. If we dissolve BaSO4 in pure water, we should get [barium]= 10^-5 and [sulfate]=10^-5. In this case, Ksp=[Ba]^2 because both ions have equal concentrations at equilibrium. But let’s imagine we have 1M NaSO4 already in solution. We add barium nitrate. Remember, Ksp=[barium][sulfate]=10^-10. In this case, because [sulfate]=1M, we only need a [barium] of 10^-10 for the Ksp to be reached. In this case Ksp≠[Ba]^2. Did I work this this out incorrectly?

I think you're confusing Ksp with Qsp. If you have Ksp = 10^-10 = [Ba][SO4] and [SO4] = 1, then that just means [SO4] isn't a limiting reagent and Qsp is dependent entirely on [Ba]. As you add more [Ba], the Qsp increases, but only up to a certain point, that point is Ksp and the max amount of dissolved [Ba] can only ever equal 1^-5. In other words, the max amount of [Ba] (aq) doesn't increase even if [SO4] >>> [Ba]. If Qsp >= Ksp, any additional [Ba(NO3)2] > 10^-5 added will contribute to BaSO4(s). In your scenario Qsp = (10^-10)^2 = 1^-20, an undersaturated solution, because only the amount of [SO4] that equals [Ba] can react and solvate [Ba].