I think Kobayashi's approach also works in the OP's situation:

If $M$ is an oriented compact connected manifold then $cd_\mathbb{R}(G) \le n$.

*Proof:* As in Kobayashi the base ring is the field of real numbers. Let $X = M \times \mathbb{R}^n$ and let $A$ be an $\mathbb{R}G$-module. Since $G$ is torsion-free and acts properly discontinuously, the action is actually free. Thus by [Cartan-Eilenberg: Homological Algebra, XVI §9] there is a spectral sequence
$$E_2^{i,j}=H^i(G;H^j(X;A)) \Rightarrow H^{i+j}(X/G;A)$$
(see also Kobayashi, p. 14). Before diving into technical details let's sketch the basic idea. Suppose $\dim M=m$ and $cd_\mathbb{R}(G) = d$. Since $X \simeq M$, $H^j(X;A)=0$ if $j>m$. Futhermore $H^i(G;-)=0$ if $i > d$. Hence the $E_2$-term looks like
$$\begin{array}{lcccr}
- & - & - & - & \bullet \newline
| & & & & | \newline
| & & & & | \newline
- & - & - & - & - \newline
\end{array}$$
Outside the rectangle all entries are zero and the bullet has coordinate $(d,m)$. For positional reasons $E_2^{d,m}=E_\infty^{d,m}$.

Now suppose $E_2^{d,m} \neq 0$. Hence the abutment $H^{d+m}(X/G;A) \neq 0$. But $X/G$ is a manifold of dimension $m+n$. Hence $H^i(X/G;A)=0$ if $i> m+n$ and consequently $d+m \le n+m$, i.e. $d \le n$ as to be shown.

The details: By $X \simeq M$ and Poincare duality we have isomorphisms $$H^m(X)\cong H^m(M) \cong H_0(M) \cong \mathbb{R}$$ as $\mathbb R$-vector spaces. Under this isomorphism the action of $g \in G$ on $H^m(X)$ corresponds to scalar multiplication on $\mathbb R$ by an $\rho(g) \in \mathbb R^\times$ satisfying $\rho(g)\rho(h)=\rho(gh)$ for $g,h \in G$. Moreover, by universal coefficients
$$H^m(X;A) \cong H^m(X) \otimes_\mathbb{R} A \cong A \hspace{110pt}(1)$$
as vector spaces and hence
$E_2^{i,m}=H^i(G;A)$.

Assume for the moment we know that the action $\odot$ of $G$ on $A$ in $(1)$ is given by
$$g \odot a = \rho(g)(g \cdot a).\hspace{150pt}(2)$$
Let $(B,\ast)$ be an $\mathbb R$-module such that $H^d(G;B) \neq 0$. Define a $G$-action $\cdot$ on $A := B$ by $$g \cdot a := \rho(g^{-1})(g\ast a)$$ (this is indeed an action because multiplication in $\mathbb R$ is commutative). Hence $\odot = \ast$ and $E_2^{d,m}=H^d(G;B) \neq 0$ as desired.

It remains to show $(2)$. This is a straightforward, but tedious calculation using the definition of the various $G$-actions. So I think it's best -- to leave it to the reader ? No, but I think it's best to write it down here only if one is really interested in the demonstation.