Reducible or irreducible? A reducible representation can be of arbitrary high dimension. An irreducible representation always has dimension $\leq\sqrt{\left|G\right|}$, if the base field is algebraically closed and of characteristic $0$ (in fact, in this case, the sum of the squares of the dimensions of all irreducible representations is $\left|G\right|$). However, if the base field is not necessarily algebraically closed and of arbitrary characteristic, then we can only say that the dimension of any irreducible representation is $\leq \left|G\right|$. Here is a proof (since I don't remember seeing this assertion in literature):
Proposition. If $G$ is a finite group, and $k$ is a field, then any irreducible representation of $G$ over $k$ has dimension $\leq \left|G\right|$.
Proof. Let $V$ be an irreducible representation of $G$ over $k$. We have to show that $\dim V\leq \left|G\right|$.
Consider the map $\rho:k\left[G\right]\to\mathrm{End}V$. This map is a map of $G$-representations, where $\mathrm{End}V$ is made a $G$-representation by $gf=\left(v\mapsto g\cdot f\left(v\right)\right)$ for any $g\in G$ and $f\in\mathrm{End}V$. Hence, its image $\rho\left(k\left[G\right]\right)$ is a $G$-subrepresentation of $\mathrm{End}V$.
But $\mathrm{End}V\cong nV$ as $G$-representations (the isomorphism is $\mathrm{End}V\to nV,\ f\mapsto\left(f\left(e_1\right),f\left(e_2\right),...,f\left(e_n\right)\right)$, where $\left(e_1,e_2,...,e_n\right)$ is any basis of the vector space $V$). Now, by Proposition 2.2 in Pavel Etingof's "Introduction to representation theory" (see Remark 2.3 for how it is independent of $k$ being algebraically closed or not), every subrepresentation of $nV$ is isomorphic to $rV$ for some nonnegative integer $V$. Since $\mathrm{End}V\cong nV$, this yields that every subrepresentation of $\mathrm{End}V$ is isomorphic to $rV$ for some nonnegative integer $r$.
Now, $\rho\left(k\left[G\right]\right)$ is a subrepresentation of $\mathrm{End}V$. Hence, $\rho\left(k\left[G\right]\right)\cong rV$ for some nonnegative integer $r$. Thus, $\dim\left(\rho\left(k\left[G\right]\right)\right)=\dim\left(rV\right)=r\dim V\geq\dim V$ (since $r\geq 1$, because otherwise $r$ would be $0$, yielding $\rho\left(k\left[G\right]\right)\cong rV=0$ and thus $\rho=0$ and consequently $V=0$, which contradicts to the condition that $V$ is an irreducible representation). But $\dim\left(\rho\left(k\left[G\right]\right)\right)=\left|G\right|$. Hence, $\left|G\right|\geq\dim V$, qed.