This is not a complete answer, but it's too long for a comment. Here is what I would try (for $G=C_p\rtimes C_2$ for any $p$): your $M$ sits in the exact sequence $$ 0\rightarrow M\rightarrow \mathbb{Z}[G/C_p] \rightarrow \boldsymbol{1} \rightarrow 0, $$ where $\boldsymbol{1}$ denotes the free $\mathbb{Z}$-rank 1 module with trivial $G$-action. This gives rise to the long exact sequence $$ 0\rightarrow M^G=0\rightarrow \boldsymbol{1}\stackrel{m}{\rightarrow}\boldsymbol{1}\rightarrow H^1(G,M)\rightarrow H^1(G,\mathbb{Z}[G/C_p])\rightarrow \\ \rightarrow H^1(G,\boldsymbol{1})\rightarrow H^2(G,M)\rightarrow H^2(G,\mathbb{Z}[G/C_p])\rightarrow \boldsymbol{1}\rightarrow \ldots $$ When you write out the map $m$, you see that it is multiplication by 2 (that's because $\mathbb{Z}[G/C_p]$ is actually indecomposable, i.e. involutions reflect this rank 2 lattice in a diagonal).
Also Shapiro's lemma gives you that $H^1(G,\mathbb{Z}[G/C_p]) \cong H^1(C_p,\boldsymbol{1})={\rm Hom}(C_p,\mathbb{Z}) = 0$, and similarly $H^1(G,\boldsymbol{1}) = 0$. Now here are some miscellaneous observations.
$H^1(G,M)\cong{\rm coker}\; m \cong \mathbb{Z}/2\mathbb{Z}$.
Explicit calculations will give you $H^2(G,\mathbb{Z}[G/C_p])\cong H^2(C_p,\boldsymbol{1})\cong \mathbb{Z}/p\mathbb{Z}$, while $H^2(G,\boldsymbol{1})\cong \mathbb{Z}/2\mathbb{Z}$, so $$ H^2(G,M)\cong {\rm ker}(\mathbb{Z}/p\mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}). $$ This is clearly equal to $\mathbb{Z}/p\mathbb{Z}$ when $p$ is odd; according to MAGMA it's also true when $p=2$, but I guess to see that you would actually have to write out the map $H^2(C_p,\boldsymbol{1})\rightarrow H^2(G,\boldsymbol{1})$.
You should be able to get quite a lot of information about higher $H^i$ if you carefully follow through the quotient map and Shapiro and write down the maps $H^i(C_p,\boldsymbol{1})\rightarrow H^i(G,\boldsymbol{1})$.
MAGMA can compute some small cohomology groups (up to $H^3$). For example here is the code for computing cohomology of this particular module:
G:=DihedralGroup(5);
M:=GModule(G, [Matrix([[1]]),Matrix([[-1]])]);
MC:=CohomologyModule(G,M);
CohomologyGroup(MC,3); //Computes $H^3$
You canMAGMA also allows you to play around with concrete co-cycles and thereby write down connecting homomorphisms between these guys.
Of course, you can do the same thing starting with the sequence $0\rightarrow \boldsymbol{1}\rightarrow \mathbb{Z}[G/C_p]\rightarrow M\rightarrow 0$.
Sorry, that's all I have to offer for now.