Sorry, I won't have time for a few days to work out the details, but I'd be happy for anybody else to do so in the meantime, Or to point out why this won't work!

For the other implications and inequalities, I think you need to consider a looser kind of filtration than the Gabriel filtration, where $\mathcal{K}_{\alpha+1}/\mathcal{K}_\alpha$ is generated by simple objects as a localizing subcategory of $\mathcal{K}/\mathcal{K}_\alpha$, but not necessarily all the available simple objects. Such a looser filtration may grow strictly more slowly than the Gabriel filtration, but should give an upper bound for the Gabriel dimension. And I think that splicing, and intersecting the Gabriel flitration of $\mathcal{K}$ with $\mathcal{T}$ does work now.

Isn't $\mathcal{K}$ Gabriel iff it has no proper quotient by a localizing subcategory that has no simple modules? In which case, it is clear that $\mathcal{K}$ Gabriel implies $\mathcal{K}/\mathcal{T}$ Gabriel (and hypoabelian groups are stable under subgroups!) But the same argument doesn't show that $\mathcal{K}$ Gabriel implies $\mathcal{T}$ Gabriel.

@SnakeEyes If it’s not clear from the construction of the stabilization of a left/right triangulated category $\mathcal{T}$, it should be clear from the universal property of the functor $\mathcal{T}\to S(\mathcal{T})$.

Do you want an exception when $A$ is self-injective? I.e., when $\underline{A}=0$? In fact, you probably want $A$ to be connected, or at least have no nonzero self-injective direct factors.

@uno To satisfy Krull-Schmidt means that every object is a finite direct sum of indecomposable objects with local endomorphism rings. But $M$ is indecomposable and doesn’t have a local endomorphism ring. See Krull-Schmidt category.

I've pretty much gone most of my career doing precisely that. But that doesn't mean that $\mathbb{Z}$ is pathological ... it just means it's HARD!

Wow! It takes a hardcore category theorist to think that $\mathbb{Z}$ is pathological!

Previously posted on MSE.

@jb2g4 The fact I mentioned is quite easy (just consider what the mapping cone of a split monomorphism of complexes looks like). A more general useful fact is that if $X\to Y$ is a map of complexes that is a split monomorphism in each degree (much weaker than being a split monomorphism of complexes), then the mapping cone is homotopy equivalent to the cokernel. I’m sure this fact can be found in many sources, but it is Exercise 11.8 in Kashiwara & Schapira’s Categories and Sheaves, for example (with hints).

@jb2g4 The cokernel of a split monomorphism is homotopy equivalent to the mapping cone, so yes, if the mapping cone is homotopy equivalent to a complex of projectives, then so will the cokernel.