For $n=2$, the answer is given by $\frac{E}{6}$, where $E$ is the Euler characteristic of $M$, see McKean Jr, H. P., & Singer, I. M. (1967). Curvature and the eigenvalues of the Laplacian. Journal of Differential Geometry, 1(1-2), 43-69.

For $n>2$, if it is topological, then it is none of what you conjectured. Consider the case of hyperbolic manifolds: since those are locally isometric to the hyperbolic space, the coefficient $b^{M}_{n/2}$ is proportional to the hyperbolic volume (Which, incidentally, is indeed topological by Mostow rigidity.)

For $n=2$, the answer is given by $\frac{E}{6}$, where $E$ is the Euler characteristic of $M$, see McKean Jr, H. P., & Singer, I. M. (1967). Curvature and the eigenvalues of the Laplacian. Journal of Differential Geometry, 1(1-2), 43-69.

For $n>2$, if it is topological, then it is none of what you conjectured. Consider the case of hyperbolic manifolds: since those are locally isometric to the hyperbolic space, the coefficient $b^{M}_{n/2}$ is proportional to the hyperbolic volume (Which, incidentally, is indeed topological by Mostow rigidity.)

Update: It is not topological for $n>2$. Let $M_1=\mathbb{T}^2\times\mathbb{S}^2$ and $M_2=(2\mathbb{T}^2)\times \mathbb{S}^2,$ where $2\mathbb{T}^2$ denotes the torus scaled by the factor of $2$. Then, $M_1$ and $M_2$ are homeomorphic and $b_2^{M_2}=4b_2^{M_1},$ because both spaces are homogeneous and locally isometric to each other. On the other hand, $$P^{M_1}_t(x,x)=P^{\mathbb{T}^2}_t(x,x)P^{\mathbb{S}^2}_t(x,x)=\left(\frac{1}{4\pi t}+o(t^{100})\right)P^{\mathbb{S}^2}_t(x,x).$$ Therefore, we only need to show that $b^{\mathbb{S}^2}_2$ does not vanish. According to the table on page 63 of the above reference, this is indeed the case.