= Solution
Sending a matrix to its last column gives the <special orthogonal sphere fibration>
$$
SO(n-1)\longrightarrow SO(n)\longrightarrow S^{n-1},\qquad n\geq2.
$$
The action is transitive and the stabilizer of that column is $SO(n-1)$. Local orthonormal frames, obtained by <Gram-Schmidt process> near a chosen unit vector, give local sections and hence local bundle trivializations. This allows the <long exact sequence of homotopy groups of a fibration> to be applied.
At the bottom of the induction, $SO(1)$ is a point and $SO(2)$ is the rotation circle $S^1$. For $SO(3)$, identify $\mathbb R^3$ with imaginary <quaternions>. A <unit quaternion> acts by $v\mapsto qvq^{-1}$, preserving length and orientation. Every rotation through angle $\theta$ about a unit imaginary axis $u$ comes from $q=\cos(\theta/2)+u\sin(\theta/2)$. The kernel is precisely $\{1,-1\}$, since a <quaternion> commuting with all imaginary <quaternions> is real. Thus this gives the twofold <covering space> $S^3\to SO(3)$. Its simply connected total space is the <universal cover>, yielding $\pi_1SO(3)=\mathbb Z/2$. Coverings induce isomorphisms on higher <homotopy groups>, so $\pi_2SO(3)=\pi_2S^3=0$ and $\pi_3SO(3)=\pi_3S^3=\mathbb Z$.
For $n\geq4$, both $\pi_1S^{n-1}$ and $\pi_2S^{n-1}$ vanish. The <Serre fibration> sequence therefore identifies $\pi_1SO(n-1)$ with $\pi_1SO(n)$ and propagates vanishing of $\pi_2$. Path-connectedness also propagates from connected fiber and base. Consequently the full low-degree answer is
$$
\boxed{\pi_0SO(n)=\{*\},\quad\pi_2SO(n)=0\quad(n\geq1),\qquad
\pi_1SO(n)=\begin{cases}0,&n=1,\\\mathbb Z,&n=2,\\\mathbb Z/2,&n\geq3.\end{cases}}
$$
Here $\pi_0$ records the single connected component.
For the remaining third group use $SO(3)\to SO(4)\to S^3$. The relevant exact segment, using the supplied fourth <homotopy group>, is
$$
\mathbb Z/2\longrightarrow\mathbb Z\longrightarrow\pi_3SO(4)\longrightarrow\mathbb Z\longrightarrow0.
$$
The first homomorphism is zero since a <torsion group> cannot map nontrivially to $\mathbb Z$. Thus it reduces to a <short exact sequence> $0\to\mathbb Z\to\pi_3SO(4)\to\mathbb Z\to0$. These groups are abelian and the final free group has a lift of its generator, so the sequence splits. Therefore
$$
\boxed{\pi_3SO(1)=\pi_3SO(2)=0,\qquad\pi_3SO(3)=\mathbb Z,\qquad\pi_3SO(4)=\mathbb Z\oplus\mathbb Z.}
$$
These are the <low homotopy groups of special orthogonal groups>.
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