Alright 3rd time's the charm - editing again to put all my cards on the table. Strap yourselves in...
For $n>2$ getting lower bounds on $c(M)$ is a difficult problem. I am interested in the particular case where $M$ is a closed, orientable, simply-connected 4-manifold. It is conjectured that $c(M)$ is bounded below by $2\chi(M)$ (ignoring additive constants). To prove this, I wonder if we can take a "local-to-global" type argument in the sense that I will explain as follows.
Fix a triangulation $\mathcal{T}$ of $M$. For a vertex $V$ of $\mathcal{T}$, consider the link of $V$: $\mathrm{lk}(V)$ -- this is the boundary of a regular neighbourhood of $V$. Suppose the $f$-vector (the vector whose $i$th entry is the number of $i$-simplices in the given complex) of $\mathrm{lk}(V)$ is $(v,e,f,t)$. Now consider the star of $V$, which is a regular neighbourhood of $V$. This can be realised as $\mathrm{st}(V)\cong C(\mathrm{lk}(V))$, where $C(K)$ is a cone over over $K$. As such,If we write the $f$-vector of $\mathrm{st}(V)$ can be written as $(v',e',f',t',p')$, then since the star is the cone of the link, we have $$(v',e',f',t',p')=(v+1,e+v,f+e,t+f,t),$$ (this can be readily seen by considering how the cone over something is constructed) where the $v,e,f,t$ come from the $f$-vector of $\mathrm{lk}(V)$.
The lower bound on $c(M)$ I desire could be shown (amongst many other equivalent formulations) if one can show that $$X(\mathcal{T})\leq 2P(\mathcal{T})+1,$$ where $P(\mathcal{T})$ is the number of pentachora in $\mathcal{T}$ (the lower bound comes from bounding the Euler characteristic of $M$, $\chi(M)=X(\mathcal{T})-T(\mathcal{T})+P(\mathcal{T})=X(\mathcal{T})-\frac{3}{2}P(\mathcal{T})$:)