Every irreducible finite-dimensional representation is either $1$-dimensional (and there are exactly $n$ of them, corresponding to the vertices) or $n$-dimensional and they can be indexed by non-zero, complex numbers.
Proof: Let $V$ be an irreducible representation, $V=\bigoplus_{i\in\mathbb{Z}/n} V_i$ its standard decomposition given by the vertices and $f_i : V_i\to V_{i+1}$ the linear maps given by the edges. Let $L:=f_{n-1}\ldots f_1 f_0$ be the endomorphism of $V_0$ given by the closed loop in the quiver.
If $V$ is finite-dimensional, then there is a eigenvector $v_0$ of $L$ for the eigenvalue $\lambda$ (we can assume that there are eigenvectors at all by assuming $V_0\neq 0$ wlog). Define $v_1:=f_0(v_0), v_2:=f_1(v_1), \ldots, v_{n-1}:=f_{n-2}(v_{n-2})$.
If $\lambda=0$, then $f_i(v_i)=0$ for some $i$ and $\mathbb{C} v_i$ is a one-dimensional submodule. By simplicity it is the whole of $V$ and we're in the first case.
If $\lambda\neq 0$, then all $v_i$ are nonzero and $\sum_i \mathbb{C}v_i$ is a submodule. By simplicity it is all of $V$, $\{v_i\}$ is a basis and one can easily write down all matrices of the edge elements w.r.t. this basis. Therefore this is a complete set of pairwise non-isomorphic irreducibles.
In fact it should be possible to describe all finite-dimensional modules in terms of a Jordan normal form of $L$. More abstractly there should be something close to a Morita equivalence between the quiver algebra $\mathbb{C}Q$ and $e_0 \mathbb{C}Q e_0$ which is just the polynomial ring $\mathbb{C}[L]$. Note that the irreducible finite-dimensional representations of $\mathbb{C}[L]$ are parametrized by complex numbers. There seems to be some sort of geometry here lurking in the background where $Irr(\mathbb{C}[L])$ is just the affine line and $Irr(\mathbb{C}Q)$ is the affine line with $n$-fold origin. Maybe some algebraic geometer can shed some light on this observation.
