Serre's finiteness theorem says if $n$ is an odd integer, then $\pi_{2n+1}(S^{n + 1})$ is the direct sum of $\mathbb{Z}$ and a finite group. By looking at the table of homotopy groups, say on Wikipedia, one empirically observes that if $n \equiv 3 \pmod 4$, then we in fact have
$$
\pi_{2n+1}(S^{n+1}) \cong \mathbb{Z} \oplus \pi_{2n}(S^n).
$$
This holds for all the cases up to $n = 19$. On the other hand, for $n \equiv 1 \pmod 4$ (and $n \neq 1$), the order of the finite part drops by a factor of $2$ when passing from $\pi_{2n}(S^n)$ to $\pi_{2n+1}(S^{n+1})$.
From the EHP sequence, we know that these two are the only possible scenarios. Indeed, we have a long exact sequence
$$
\pi_{2n+2}(S^{2n+1}) \cong \mathbb{Z}/2\overset{P}{\to} \pi_{2n}(S^n) \overset{E}{\to} \pi_{2n+1}(S^{n+1}) \overset{H}{\to}\pi_{2n+1}(S^{2n+1}) \cong \mathbb{Z}.
$$
Since $H$ kills of all torsion, one sees that the map $E$ necessarily surjects onto the finite part of $\pi_{2n+1}(S^{n+1})$. So the two cases boil down to whether or not $P$ is the zero map. What we observed was that it is zero iff $n \equiv 3 \pmod 4$, up to $n = 19$.
Since $\pi_{2n+2}(S^{2n+1}) \cong \mathbb{Z}/2\mathbb{Z}$ has only one non-zero element, which is the suspension of the Hopf map, it seems like perhaps one might be able to check what happens to this element directly. However, without a good grasp of what the map $P$ (or the preceeding $H\colon\pi_{2n+2}(S^{n+1}) \to \pi_{2n+2}(S^{2n+1})$) does, I'm unable to proceed.
Curiously, I can't seem to find any mention of this phenomenon anywhere. The closest I can find is this MO question, but this phenomenon is not really about early stabilization, since for $n = 3, 7$, the group $\pi_{2n-1}(S^n)$ is not the stable homotopy group, but something smaller. I'd imagine either this pattern no longer holds for larger $n$, or there is some straightforward proof I'm missing.
Note: Suggestions for a more descriptive title are welcome.
Best Answer
$P(\eta) = [i_n,i_n] \circ \eta$, where $[i_n,i_n]: S^{2n-1} \rightarrow S^n$ is the Whitehead product of the identity map with itself. So you are asking if this composite is null.
I don't know if this is an easy problem. One sufficient condition is that $[i_n,i_n]$ be divisible by 2, but, ha, ha, this is now known to only rarely happen, thanks to the Hill-Hopkins-Ravenel theorem on the Kervaire invariant. (See [HHR, Thm 1.5].) But it does for $n=63$, so $P$ is zero in that case.
I'd search the old literature - papers of Mahowald, Barratt, James and their collaborators - for any general results, if they exist. Mahowald, in particular, has many papers with examples of families of elements on which $H$ acts nonzero.
Added 30 minutes later: [Mahowald, Some Whitehead products in $S^n$, Topology 4, 1965, Theorem 1.1.2(a)] answers your question. $P(\eta)$ is zero if $n \equiv 3 \mod 4$ and is nonzero in basically all other cases.