EDIT: Part 4 added.
1. The answer is no (as long as we are working over a field - of any characteristic, algebraically closed or not). If $k$ is a field and $G$ is a finite group, then the dimension of any irreducible representation $V$ of $G$ over $k$ is $\leq \left|G\right|$. This is actually obvious: Take any nonzero vector $v\in V$; then, $k\left[G\right]v$ is a nontrivial subrepresentation of $V$ of dimension $\leq\dim\left(k\left[G\right]\right)=\left|G\right|$. Since our representation $V$ was irreducible, this subrepresentation must be $V$, and hence $\dim V\leq\left|G\right|$.
2. Okay, we can do a little bit better: Any irreducible representation $V$ of $G$ has dimension $\leq\left|G\right|-1$, unless $G$ is the trivial group. Same proof applies, with one additional step:
If $\dim V=\left|G\right|$, then the map $k\left[G\right]\to V,\ g\mapsto gv$ must be bijective (in fact, it is surjective, since $k\left[G\right]v=V$, and it therefore must be bijective since $\dim\left(k\left[G\right]\right)=\left|G\right|=\dim V$), so it is an isomorphism of representations (since it is $G$-equivariant), and thus $V\cong k\left[G\right]$. But $k\left[G\right]$ is not an irreducible representation, unless $G$ is the trivial group (in fact, it always contains the $1$-dimensional trivial representation).
3. Note that if the base field $k$ is algebraically closed and of characteristic $0$, then we can do much better: In this case, an irreducible representation of $G$ always has dimension $<\sqrt{\left|G\right|}$ (in fact, in this case, the sum of the squares of the dimensions of all irreducible representations is $\left|G\right|$, and one of these representations is the trivial $1$-dimensional one). However, if the base field is not necessarily algebraically closed and of arbitrary characteristic, then the bound $\dim V\leq \left|G\right|-1$ can be sharp (take cyclic groups).
4. There is a way to improve 2. so that it comes a bit closer to 3.:
If $V_1$, $V_2$, ..., $V_m$ are $m$ pairwise nonisomorphic irreducible representations of a finite-dimensional algebra $A$ over a field $k$ (not necessarily algebraically closed, not necessarily of characteristic $0$), then $\dim V_1+\dim V_2+...+\dim V_m\leq\dim A$.
(Of course, if $A$ is the group algebra of some finite group $G$, then $\dim A=\left|G\right|$, and we get 2. as a consequence.)
Proof. At first, for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, the (left) representation $V_i^{\ast}$ of the algebra $A^{\mathrm{op}}$ (this representation is defined by $a\cdot f=\left(v\mapsto f\left(av\right)\right)$ for any $f\in V_i^{\ast}$ and $a\in A$) is irreducible (since $V_i$ is irreducible) and therefore isomorphic to a quotient of the regular (left) representation $A^{\mathrm{op}}$ (since we can choose some nonzero $u\in V_i^{\ast}$, and then the map $A^{\mathrm{op}}\to V_i^{\ast}$ given by $a\mapsto au$ must be surjective, because its image is a nonzero subrepresentation of $V_i^{\ast}$ and therefore equal to $V_i^{\ast}$ due to the irreducibility of $V_i^{\ast}$). Hence, by duality, $V_i$ is isomorphic to a subrepresentation of the (left) representation $A^{\mathrm{op}\ast}=A^{\ast}$ of $A$. Hence, from now on, let's assume that $V_i$ actually is a subrepresentation of $A^{\ast}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$.
Now, let us prove that the vector subspaces $V_1$, $V_2$, ..., $V_m$ of $A^{\ast}$ are linearly disjoint, i. e., that the sum $V_1+V_2+...+V_m$ is actually a direct sum. We will prove this by induction over $m$, so let's assume that the sum $V_1+V_2+...+V_{m-1}$ is already a direct sum. It remains to prove that $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)=0$. In fact, assume the contrary. Then, $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)=V_m$ (since $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)$ is a nonzero subrepresentation of $V_m$, and $V_m$ is irreducible). Thus, $V_m\subseteq V_1+V_2+...+V_{m-1}$. Consequently, $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ (because the sum $V_1+V_2+...+V_{m-1}$ is a direct sum, according to our induction assumption).
Now, according to Theorem 2.2 and Remark 2.3 of Etingof's "Introduction to representation theory", any subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ must be a direct sum of the form $r_1V_1\oplus r_2V_2\oplus ...\oplus r_{m-1}V_{m-1}$ for some nonnegative integers $r_1$, $r_2$, ..., $r_{m-1}$. Hence, every irreducible subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ must be one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. Since we know that $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$, we conclude that $V_m$ is isomorphic to one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. This contradicts the non-isomorphy of the representations $V_1$, $V_2$, ..., $V_m$. Thus, we have proven that the sum $V_1+V_2+...+V_m$ is actually a direct sum. Consequently, $\dim V_1+\dim V_2+...+\dim V_m=\dim\left(V_1+V_2+...+V_m\right)\leq \dim A^{\ast}=\dim A$, qed.