The question can be considered and answered in greater generality: 

> Let $R$ be a (not necessarily commutative) ring with unit and let $G$ be a finite group. Then the trivial $RG$-module $R$ is flat iff $|G|$ is invertible in $R$.

Proof: By a result of Benson, $R$ is flat iff it's projective (see [Theorem 1.2][1]), which is equivalent to the splitting of the augementation $\epsilon: RG \to R$ (over $RG$). 

Assume $i: R \to RG$ is a spliting of $\epsilon$ and $i(1) = \alpha$. From $g \cdot i(1) = i(g\cdot 1) = i(1)\;\; (g \in G)$ it follows that $\alpha = r \cdot N_G$ for some $r \in R$ and the norm element $N_G = \sum_{g \in G}g$. Appyling $\epsilon$ yields $1=r\cdot |G|$, so $|G|$ is a unit in $R$. 

Conversely, if $|G|$ is a unit in $R$, than $i:R \to RG, 1 \to |G|^{-1}\cdot N_G$ is easily seen to be a splitting of $\epsilon$. 

  [1]: http://www.maths.abdn.ac.uk/~bensondj/html/archive/benson-goodearl.html