The question can be 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), 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 $\epsilon$.