There is no need of representation theory. The orthogonal group is the set of extremal points of the unit sphere associated with the standard (operator) norm over $M_n({\mathbb R})$. Therefore it spans the same subspace as the unit sphere: the entire space.
Edit. Since this question was about $SO_n$ from the begining, here is an appropriate answer: the convex hull of $SO_n$ has been described by J. Saunderson, P. A. Parrilo, and A. S. Willsky in this paper. It can be used to prove that $SO_n$ spans the whole of $M_n({\mathbb R})$.
Elementary proof. The linear space $E$ spanned by $SO_n$ is the orthogonal of those matrices $M$ such that $\langle M,Q\rangle:={\rm Tr}(MQ)=0$ for every $Q\in SO_n$. Let $M=SR$ be a polar decomposition, where $S\in Sym_n^+$ and $R\in O_n$. This decomposition is unique with $S\in SPD_n$ if $M$ is non-singular, but in general it exists and might be non-unique. If $R\in SO_n$, then ${\rm Tr}(SQ)=0$ for every $Q\in SO_n$ ; chosing $Q=I_n$, we have ${\rm tr}\,S=0$, which implies $S=0_n$. If on the contrary $R\in O_n^-$, we have ${\rm Tr}(SQ)=0$ for every $Q\in O_n^-$. Diagonalize $S$ in an orthogonal basis, the matrix of $D$ eigenvalues satisfies ${\rm Tr}(DQ)=0$ for every $Q\in O_n^-$. Chosing $Q$ the symmetry with respect to hyperplane $x_j=0$, we obtain ${\rm Tr}\,D=2d_j$ for every $j$. There follows $n\,{\rm Tr}\,D=2\,{\rm Tr}\,D$. Whence ${\rm Tr}\,D=0$, $d_j=0$ if $n\ge3$. This yields $M=0_n$, hence $E$ is the full space $M_n$.
When $n=2$, this gives only $d_1=d_2$ and we recover $M\in O_2^-$.