Why does [Os2Cl8]2- not form a quadruple bond?

Well, like you've said, you have 5 electrons contributed per Os and that means that if you had an eclipsed Cl₄Os–OsCl₄ geomtery, δ* would be filled. In other words, you'd have a σ²π⁴δ²δ*² configuration. Bond order = (8 – 2)/2 = 3. The two antibonding electrons cancel the δ bond and the bond order goes down to a triple bond. The geometry is consequently staggered, not eclipsed. In the eclipsed geometry, the d-d δ overlap is eliminated; instead, each of the dδ orbitals (on each square-pyramidal OsCl₄ fragment) only overlaps with the highly destabilized d orbital on the other fragment that is pushed up by interaction with the chloride ligand donor orbitals. Basically then, in the staggered geometry, those two δ-symmetry d orbitals remain non-bonding.

EDIT: Apparently the Os₂X₈²⁻ ions USUALLY adopt staggered conformations, but not always (see the 2nd link below). No matter, it should still have only a triple bond, as discussed above.