Building on the results established by Tom and Tyler, it seems to me that the structure of $H_\ast(Sym^n M)$ can be further explored:

Since the characteristic of $k$ is assumed to be zero, $H_\ast(Sym^n M) = H_\ast(M^{\otimes n})_{\Sigma_n}$ by Tom's and Tyler's answer and by Künneth formula
$H_\ast(M^{\otimes n}) = H_\ast(M)^{\otimes n}$. Hence as an intermediate result

$$H_\ast(Sym^n M) = H_\ast(M)^{\otimes n}_{\Sigma_n}$$

Consider $H_\ast(M)$ as a graded $k$-vector space and choose a basis $\lbrace x_j \mid j \in J\rbrace$ where $J$ is assumed to be totally ordered (denote the order by $\le$). Then the elements $x_{j_1} \otimes \cdots \otimes x_{j_n}=: \underline{x}_j$, $\;j = (j_1,...,j_n) \in J^n$ form a basis of $H_\ast(M)^{\otimes n}$ and the action of $\Sigma_n$ is given by
$$\sigma^{-1} \cdot x_{j_1} \otimes \cdots \otimes x_{j_n} = x_{i_1} \otimes \cdots \otimes x_{i_n}$$
where
$$(i_1,...,i_n) = (\sigma(j_1),...,\sigma(j_n))=: \sigma^{-1} \cdot (j_1,...,j_n).$$

The latter defines an action of $\Sigma_n$ on $J^n$ as well. We can subdivide $J^n$ into those tuples having exactly $l$ equal elements $(l= 1,...,n)$. This leads to the following set of representatives $E = J^n/\Sigma_n = \coprod_{l=1}^n\; E_l$ where

$$ E_1 = \lbrace (j_1,...,j_n) \mid j_1 < ... < j_n\rbrace $$
$$ E_l = \lbrace (j,...,j,j_1,...,j_{n-l}) \mid j_1 < ... < j_{n-l},\;\; j \neq j_i\; \forall i\rbrace\quad(2 \le l \le n)$$

The stabilizer of $j \in E_l$ is $\Sigma_l \le \Sigma_n$. Now by Corollary III.5.4 in Brown, Cohomology of Groups:

$$H_\ast(M)^{\otimes n} = \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; \operatorname{Ind}^{\Sigma_n}_{Stab(j)}k\cdot \underline{x}_j$$

$$\hspace{60pt} = \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; k[\Sigma_n] \otimes_{k[\Sigma_l]}\;k\cdot \underline{x}_j$$

Hence by Brown, II (2.1):

$\qquad H_\ast(M)^{\otimes n}_{\Sigma_n} = k \otimes_{k[\Sigma_n]} H_\ast(M)^{\otimes n}$ $= \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; k \otimes_{k[\Sigma_l]} \;k\cdot \underline{x}_j$

$\hspace{60pt}= \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; k\cdot \underline{x}_j$

So as final result we keep hold:

$\qquad\qquad H_\ast(Sym^n M) = \bigoplus_{j \in E}\; k\cdot \underline{x}_j\quad, \qquad E= J^n/\Sigma_n$

In particular, if $m := \dim_k H_\ast(M) < \infty$ then

$\dim_k H_\ast(Sym^n M) = |E| = \binom{m}{n} + \sum_{l=0}^{n-2} \;\binom{m-1}{l}\;m$

For example, if $n > m$ then $\dim_k H_\ast(Sym^n M) = 2^{m-1}\;m$.

**Remark:** The question requests that there is at most one $r$ such that $N_1 := H_r(M), N_2 := H_{r+1}(M) \neq 0$. Using a (more or less obvious) result of Dold: Homology of Symmetric Products and other Functors of Complexes (1958), (8.4), the structure of $H_\ast(M)^{\otimes n}_{\Sigma_n}$ can be refined as
$$H_\ast(M)^{\otimes n}_{\Sigma_n} = \bigoplus_{p+q=n} (N_1)_{\Sigma_p} \otimes_k (N_2) _{\Sigma_q}$$
Then each $(N_i)_{\Sigma_p}$ can be separately analyzed as before.

`$C_*(EG)$`

is levelwise free over $\mathbb{Z}[G]$, this "derived coinvariants" is just the derived tensor product of $\mathbb{Z}$; any reference on derived tensor products or "hypertor" should show that it preserves weak equivalences. So far as the reference to McCleary below, I was more thinking of it as a reference for derived tensor product of modules over a DGA in order to take powers. (Topologists do work a lot with symmetric powers, but it is usually the "homotopy" symmetric power. There are a lot of sticky issues related to it which you may not want to know.) – Tyler Lawson Mar 13 '12 at 17:03