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This is not a complete answer in any sense, but I will make a few comments. The irreducible subgroups $G$ of ${\rm GL}(2,\mathbb{C})$ are the primitive ones, which have $G/Z(G)$ isomorphic to $A_{4},S_{4}$ or $A_{5}$, and imprimitive groups, which have an Abelian normal subgroup of index $2.$ On the other hand, any finite group with an Abelian normal subgroup of index $2$ has all its irreducible representations of degree $1$ or $2,$ so the number of $2$-dimensional irreducible representations is easily calculated. A more careful analysis of the primitive case shows that if $G$ has a faithful $2$-dimensional primitive complex representations, then $G = Z(G)E,$ where $E \cong {\rm SL}(2,3), {\rm GL}(2,3),{\rm SL}(2,5)$ or the binary icosahedral group (also, a double cover of order $48$ of $S_{4},$ (as ${\rm GL}(2,3)$ is), but with a generalized quaternion Sylow $2$-subgroup). Now let $G$ be any finite group, and let $K$ be the intersection of the kernels of the irreducble complex representation of $G$ of degree at most $2$. The above discussion means that the only possible non-Abelian composition factor of $G/K$ is $A_{5}$, though it may be repeated if it appears. The answer to your question only depends on the structure of $G/K,$ so we may reduce to the case that all composition factors of $G$ are cyclic or $A_{5}.$ Also, by Clifford theory, we may suppose that the Fitting subgroup $F(G)$ is a direct product of an Abelian group of odd order and a $2$-group, and that all components of $G$ (if there are any) are isomorphic to ${\rm SL}(2,5).$ Further analysis can be carried out, but I believe that the analysis is not entirely straightforward. Perhaps this outline will help others to complete it, so I submit it.
This is not a complete answer in any sense, but I will make a few comments. The irreducible subgroups $G$ of ${\rm GL}(2,\mathbb{C})$ are the primitive ones, which have $G/Z(G)$ isomorphic to $A_{4}$ A_{4},S_{4}$or$A_{5}$, and imprimitive groups, which have an Abelian normal subgroup of index$2.$On the other hand, any finite group with an Abelian normal subgroup of index$2$has all its irreducible representations of degree$1$or$2,$so the number of$2$-dimensional of$2$-dimensional irreducible representations is easily calculated. A more careful analysis of the primitive case shows that if$G$has a faithful$2$-dimensional primitive complex representations, then$G = Z(G)E,$where$ E \cong {\rm SL}(2,3)$SL}(2,3), {\rm GL}(2,3),{\rm SL}(2,5)$ or the binary icosahedral group (also, a double cover of order $48$ of $S_{4},$ (as ${\rm SL}(2,5).$ GL}(2,3)$is), but with a generalized quaternion Sylow$2$-subgroup). Now let$G$be any finite group, and let$K$be the intersection of the kernels of the irreducble complex representation of$G$of degree at most$2$. The above discussion means that the only non-Abelian composition factor of$G/K$is$A_{5}$, though it may be repeated. The answer to your question only depends on the structure of$G/K,$so we may reduce to the case that all composition factors of$G$are cyclic or$A_{5}.$Also, by Clifford theory, we may suppose that the Fitting subgroup$F(G)$is a direct product of an Abelian group of odd order and a$2$-group, and that all components of$G$(if there are any) are isomorphic to${\rm SL}(2,5).$Further analysis can be carried out, but I believe that the analysis is not entirely straightforward. Perhaps this outline will help others to complete it, so I submit it. 2 tidying up This is not a complete answer in any sense, but I will make a few comments. The irreducible subgroups$G$of${\rm GL}(2,\mathbb{C})$are the primitive ones, which have$G/Z(G)$isomorphic to$A_{4}$or$A_{5}$, and imprimitive groups, which have an Abelian normal subgroup of index$2.$On the other hand, any finite group with an Abelian normal subgroup of index$2$has all its irreducible representations of degree$1$or$2,$so the number of$2$-dimensional irreducible representations is easily calculated. A more careful analysis of the primitive case shows that if$G$has a faithful$2$-dimensional primitive complex representations, then$G = Z(G)E,$where$ E \cong {\rm SL}(2,3)$or${\rm SL}(2,5).$Nowlet Now let$G$be any finite group, and let$K$be the intersection of the kernels of the irreducble complex representation of$$G$ of degree at most $2$. The above discussion means that the only non-Abelian composition factor of $G/K$ is $A_{5}$, though it may be repeated. The answer to your question only depends on the structure of $G/K,$ so we may reduce to the case that all composition factors of $G$ are cyclic or $A_{5}.$ Also, by Clifford theory, we may suppose that the Fitting subgroup $F(G)$ is a direct product of an Abelian group of odd order and a $2$-group, and that all components of $G$ (if there are any) are isomorphic to ${\rm SL}(2,5).$ Further analysis can be carried out, but I believe that the analysis is not entirely straightforward. Perhaps this outline will help others to complete it, so I submit it.