I've been examining the simple modules of group algebras over fields over than $\mathbb{C}$, and I think I have shown that any simple $\mathbb{Q}C_{n}$ module has even dimension as a $\mathbb{Q}$-vector space.
Here $C_{n}$ is the cyclic group of order n, and $\mathbb{Q}C_{n}$ denotes the group $\mathbb{Q}-$algebra of $C_{n}$.
Consider first the case when $n=p$ is prime.
Let $G = C_{p} = <\xi> $ be the cyclic group of prime order $p$. Suppose that $\rho : G \rightarrow \text{GL}(V) $ is an irreducible $\mathbb{Q}$-representation of $G$, with $V$ a finite dimensional $\mathbb{Q}$-vector space of dimension yet to be determined. Let $T = \rho(\xi)$ be the image of $ \xi$ in $\text{GL}(V)$, that $T$ is an endomorphism of $V$. Then $\xi^{p} = 1$ in $G$ and so $T^{p} = \rho(\xi)^{p} = \rho(\xi^{p}) = \rho(1) = \text{Id}_{n}$, where $n = \text{dim}_{\mathbb{Q}}(V)$. Hence $T^{p}$ is the identity map on $V$, and so the minimum polynomial $m_{T}(x)$ of $T$ divides $x^{p} - 1$. We know that, since $p$ is prime, over $\mathbb{Q}$, $x^{p} - 1 = (x - 1) \cdot \Phi_{p}(x)$ is a factorisation into irreducible polynomials, with $\Phi_{p}(x) = \sum_{i=0}^{p-1}x^{i}$ the $p^\text{th}$ cyclotomic polynomial. Then by the primary decomposition theorem, $V \cong \text{ker}((x-1)) \oplus \text{ker}(\Phi_{p}(x))$ as $\mathbb{Q}$ vector spaces.
So we can conclude that the only irreducible $\mathbb{Q}$-representations of $G$ are the trivial representation coming from the $(x-1)$ factor of $m_{T}(x)$, and a $(p-1)$-dimensional representation coming from the $\Phi_{p}$ factor since $\text{deg}(\Phi_{p}) = p-1$. Then with respect to some basis of $V$, $T$ will have a matrix $M_{T}$ say, that has minimum polynomial $m_{T}(x)$. We can use the companion matrix of $m_{T}(x)$ to find such an $M_{T}$:
$$ M_{T} = \left. \begin{array}{l} \begin{bmatrix} 0 & 1 & 0 & 0 & \dots & 0 & 0 \\ 0 & 0 & 1 & 0 & \dots & 0 & 0 \\ 0 & 0 & 0 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \dots & 1 & 0 \\ 0 & 0 & 0 & 0 & \dots & 0 & 1 \\ -1 & -1 & -1 & -1 & \dots & -1 & -1 \end{bmatrix} \end{array} \right\} (p-1) $$
Then we have shown that group $\mathbb{Q}$-algebra $\mathbb{Q}C_{p}$ has a unique (up to isomorphism) non-trivial simple module $S$ say with $\text{dim}_{\mathbb{Q}}S = p-1$, where $S$ is the $(p-1)$-dimensional $\mathbb{Q}$ vector space where the action of $\mathbb{Q}C_{p}$ on $S$ is given by $\xi \cdot \mathbf{v} = M_{T}\mathbf{v}$ for vectors $\mathbf{v} \in S$
Then, I think the above idea has an extension to when n is not prime. Since $\mathbb{Q}[x]$ is a unique factorisation domain, and $(x^{n} - 1) = \prod_{d| n}\Phi_{d}(x)$, with $\Phi_{d}(x)$ all irreducible in $\mathbb{Q}[x]$, we see that all $\mathbb{Q}$-irreducible factors of $(x^{n} - 1)$ are in fact cyclotomic polynomials. Since all cyclotomic polynomials, other than $\Phi_{1}(x) = (x-1)$ and $\Phi_{2}(x) = (x+1)$ have even degree, using the exact same construction, can we conclude that any simple $\mathbb{Q}C_{n}$-module is either the trivial module, from a $(x-1)$ factor, the sign module from a $(x+1)$ factor, or has even dimension as a $\mathbb{Q}$-vector space from a cyclotomic factor?
The group algebra $\mathbb{Q}[C_n]$ is isomorphic to the quotient ring $\mathbb{Q}[x]/(x^n-1)$. Since the cyclotomic polynomials are irreducible and coprime, by the Chinese remainder theorem the ring is isomorphic to $\bigoplus_{d\mid n} \mathbb{Q}[x]/(\Phi_d(x))$, with each summand a simple $\mathbb{Q}[C_n]$-module. Except for $d=1$ and $d=2$, these modules have even dimension. By semisimplicity, these are all the simple modules.