This should be a comment, mostly because I have not checked...

Fix $n$ and consider only $m=1$. For a natural number $\ell$ let $[\ell]=\lbrace 1,\dots,\ell\rbrace$.

Let $X_k(n)$, for each $k\geq0$, be the vector space with basis the set of all functions $[n]\to[k+1]$. Then $X_\flat(n)$ is a simplicial vector space (with simplicial operations acting on $[k+1]$). We can do better: for each $k\geq0$, there is an endomorphism $t_k:X_k(n)\to X_k(n)$ such that $t_k^{k+1}=\mathrm{id}$, given simply by rotation in $[k+1]$. The map $t_k$ plays nicely with the simplicial structure and I think that in this way $X_\flat(n)$ becomes a cyclic vector space (a cyclic object in the category of vector spaces) in the sense of Alain Connes; see Jean-Louis Loday's book on cyclic homology.

Now if the underlying field is of characteristic zero, the Connes' complex $C^\lambda$ corresponding to the cyclic vector space $X_\flat(n)$ has $C^\lambda_k=X_k(n)/(1-t_k)$; i other words, we identify things with their image under $t_k$. Elements of this vector space are functions $[n]\to[k+1]$ up to a rotation in $[k+1]$, which is the same thing as a partition of $[n]$ in $k+1$ labeled parts, up to a cyclic rotation of the labeling. There are $S(n,k+1)(k+1)!$ labeled partitions of $[n]$ into $k+1$ parts, and when we look at them up to rotation in $[k+1]$ we get $S(n,k+1) k!$. Now $c(k+1,1)=k!$, so in the end $$\dim C^\lambda_k=S(n,k+1)c(k+1,1).$$

It follows that you are almost trying to compute the Euler characeristic of the complex $C^\lambda$.

I don't have paper at hand (and this margin &c) so I cannot try now: but one knows that the homology of $C^\lambda$ can be obtained from a different complex, Connes' cyclic bicomplex or his triangular complex. Maybe the homology of those complexes can be readily computed, and that would do pretty much what you want.

This should be a comment, mostly because I have not checked...

Fix $n$ and consider only $m=1$. For a natural number $\ell$ let $[\ell]=\lbrace 1,\dots,\ell\rbrace$.

Let $X_k(n)$, for each $k\geq0$, be the vector space with basis the set of all functions surjective $[n]\to[k+1]$. Then $X_\flat(n)$ is a simplicial vector space (with simplicial operations acting on $[k+1]$). We can do better: for each $k\geq0$, there is an endomorphism $t_k:X_k(n)\to X_k(n)$ such that $t_k^{k+1}=\mathrm{id}$, given simply by rotation in $[k+1]$. The map $t_k$ plays nicely with the simplicial structure and I think that in this way $X_\flat(n)$ becomes a cyclic vector space (a cyclic object in the category of vector spaces) in the sense of Alain Connes; see Jean-Louis Loday's book on cyclic homology.

Now if the underlying field is of characteristic zero, the Connes' complex $C^\lambda$ corresponding to the cyclic vector space $X_\flat(n)$ has $C^\lambda_k=X_k(n)/(1-t_k)$; i other words, we identify things with their image under $t_k$. Elements of this vector space are functions $[n]\to[k+1]$ up to a rotation in $[k+1]$, which is the same thing as a partition of $[n]$ in $k+1$ labeled parts, up to a cyclic rotation of the labeling. There are $S(n,k+1)(k+1)!$ labeled partitions of $[n]$ into $k+1$ parts, and when we look at them up to rotation in $[k+1]$ we get $S(n,k+1) k!$. Now $c(k+1,1)=k!$, so in the end $$\dim C^\lambda_k=S(n,k+1)c(k+1,1).$$

It follows that you are almost trying to compute the Euler characeristic of the complex $C^\lambda$.

I don't have paper at hand (and this margin &c) so I cannot try now: but one knows that the homology of $C^\lambda$ can be obtained from a different complex, Connes' cyclic bicomplex or his triangular complex. Maybe the homology of those complexes can be readily computed, and that would do pretty much what you want.