Since it's not clear, I assume the question you're asking is the following. "Fix an $n\geq2$. Which finite groups have a faithful real $n$-dimensional representation? Equivalently, what are the finite subgroups of $\operatorname{GL}_n \mathbb R$?" (If you just want to know which finite groups admit faithful real representations then the answer is easy: all of them.)
Unfortunately, it's hopelessly difficult to answer this question for general $n$. See the answers of Richard Borcherds and Geoff Robinson here for an explanation.
For $n=2$, though, there is a simple answer. It turns out that a finite subgroup of $\operatorname{GL}_2\mathbb R$ is either cyclic or dihedral. Here's a sketch of the argument (for a slightly expanded version, see Rees, Notes on Geometry). First note that such a subgroup $G$ can be conjugated into the orthogonal group $\operatorname{O}(2)$ (just average an inner product over $G$). Thus we may as well assume that $G \subset \operatorname{O}(2)$. Now consider the subgroup $H = G \cap \operatorname{SO}(2)$, which consists of rotations. It's not too hard to show that $H$ is in fact generated by one of these rotations -- namely the one whose rotation angle is the smallest. Thus if $H = G$, then $G$ is cyclic and we're done. Otherwise $[G : H] = 2$ and $G \backslash H \subset \operatorname{O}(2) \backslash \operatorname{SO}(2)$ consists of reflections. This forces $G$ to be dihedral.
There is also a relatively nice answer for $n=3$, but the list of possible subgroups this time is bigger...
Edit: You can find a list of the finite subgroups of $\operatorname{O}(3)$ (hence of $\operatorname{GL}_3 \mathbb R$) here. Let me briefly indicate how you get this list. As in the case of $\operatorname{GL}_2 \mathbb R$, you start off by determining the finite subgroups of $\operatorname{SO}(3)$. These are given by Neil in his comment below. Next you use the fact that $\operatorname{O}(3) = \operatorname{SO}(3) \times \{\pm I\}$ together with Goursat's lemma to get the complete list.
