Suppose $R$ is a polynomial ring of dimension $n$ and let $M$ be a finitely generated module of projective dimension $m-1$. Suppose we have a resolution $$ 0\rightarrow F_m\rightarrow F_{m-1}\rightarrow\cdots\rightarrow F_1\rightarrow F_0\rightarrow M\rightarrow 0.$$
We know that $F_i$'s are Cohen-Macaulay ($\dim F_i=\operatorname{depth} F_i=n$). Now let $C=\mathrm{coker}(F_m\rightarrow F_{m-1})$. Is it true that $C$ is a Cohen-Macaulay module of dimension $n$?
If not, can we add some conditions which make $C$ Cohen-Macaulay of dimension $n$?
You seem to have changed your question. Let me answer your original question when $m = 1$ negatively. For instance, take $R = k[x]$ and $M = R/xR$. Then $\dim M=0$.
For simplicity, let me replace the polynomial ring by a regular local ring of dimension $n$. The Auslander-Buchsbaum formula says that $M$ is Cohen-Macaulay of dimension $n$ if and only if $M$ is free. In other words, one cannot expect $M$ to be dimension $n$ if the given resolution is minimal. If you are a geometer, look at Horrock’s formula.