The orthogonal group $O(n)$ has two path-connected components, one of which is $SO(n)$ (namely, the path-connected component containing $I$). As $O(n)$ is a Lie group, its path-connected components are diffeomorphic, so the fundamental group of each component is isomorphic to the fundamental group of $SO(n)$. Therefore, even though $O(n)$ is not path-connected, its fundamental group is well-defined, even without specifying basepoints. In particular, we have $\pi_1(O(n)) = \pi_1(SO(n))$.
For $n = 1$, we have $SO(1) = \ast$, a point, so $\pi_1(O(1)) = \pi_1(SO(1)) = 0$. As $SO(1)$ is contractible, the Whitehead tower of $O(1)$ is $\ast \to O(1)$.
For $n = 2$, we have $SO(2) = S^1$ so $\pi_1(O(2)) = \pi_1(SO(2)) = \mathbb{Z}$. The group $\operatorname{Spin}(2)$ is $SO(2)$ itself but the map $\operatorname{Spin}(2) \to SO(2)$ is the double-sheeted covering map given by $A \mapsto A^2$; in particular, it has fundamental group $\pi_1(\operatorname{Spin}(2)) = \mathbb{Z}$. As the higher homotopy groups of $SO(2)$ are trivial, its $2$-connected cover is contractible so the Whitehead tower of $O(2)$ is $* \to SO(2) \to O(2)$. In this case, $\operatorname{Spin}(2)$ isn't part of the Whitehead tower because it has the same connectivity as $SO(2)$.
I'm not sure whether $\operatorname{String}(n)$ and $\operatorname{Fivebrane}(n)$ are defined for $n = 1, 2$, but by any reasonable definition they should be contractible and hence simply connected.
Now that we have dealt with the low dimensional cases, we can treat the general case: $n \geq 3$.
First note that $\pi_1(SO(n)) \cong \mathbb{Z}_2$. The group $\operatorname{Spin}(n)$ is the universal cover (a double cover) of $SO(n)$, so $\pi_1(\operatorname{Spin}(n)) = 0$; i.e. $\operatorname{Spin}(n)$ is the $1$-connected cover of $SO(n)$, and hence of $O(n)$. Climbing the Whitehead tower, the next space is the $2$-connected cover, but it turns out that $\pi_2(\operatorname{Spin}(n)) = 0$ (the second homotopy group of a Lie group is always zero), so in fact $\operatorname{Spin}(n)$ is also the $2$-connected cover. As $\operatorname{Spin}(n)$ is a non-contractible Lie group, $\pi_3(\operatorname{Spin}(n)) \neq 0$ (in fact, because $\operatorname{Spin}(n)$ is simple, $\pi_3(\operatorname{Spin}(n)) \cong \mathbb{Z}$, so $\operatorname{Spin}(n)$ is not $3$-connected. The $3$-connected cover is denoted $\operatorname{String}(n)$. I don't know much about $\operatorname{Fivebrane}(n)$, but it is the next space up in the Whitehead tower, in particular, it is at least $4$-connected, but it might be more: $\operatorname{String}(n)$ could be more than just $3$-connected, just as $\operatorname{Spin}(n)$ was more than just $1$-connected.
This last paragraph is a bit of overkill in regards to your question. In short, we have (for $n \geq 3$) the Whitehead tower
$$\dots \to \operatorname{Fivebrane}(n) \to \operatorname{String}(n) \to \operatorname{Spin}(n) \to SO(n) \to O(n).$$
As connectivity increases as you move up the Whitehead tower, every space in the Whitehead tower after the second space is necessarily simply connected. In particular,
$$\pi_1(\operatorname{Spin}(n)) = 0$$
$$\pi_1(\operatorname{String}(n)) = 0$$
$$\pi_1(\operatorname{Fivebrane}(n)) = 0.$$
