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If this was the case, then $\pi_{n-2}(SO(n))\cong \pi_{n-2}(SO(n-1))$. The first group is in the stable range and hence one of $\mathbb{Z},\mathbb{Z}/2$ or 0; in particular every surjective endomorphism of it is an isomorphism. Now the orthonormal frame bundle of $S^{n-1}$ is a fiber sequence $SO(n-1)\to SO(n) \to S^{n-1}$ which gives a long exact sequence $$ \pi_{n-1}(SO(n))\to\pi_{n-1}(S^{n-1})\to \pi_{n-2}(SO(n-1))\to \pi_{n-2}(SO(n))\to\pi_{n-2}(S^{n-1}). $$ The last group is zero, so the third arrow is a surjection; by what was said before, it is an isomorphism. Hence the first map is surjective, and a preimage of the canonical generator gives a section of the frame bundle, in other words, a trivialization of the tangent bundle of $S^{n-1}$. But it is well-known (see user51223's answer) that this is only possible in the dimensions you have listed above.