I agree. Just wanted to make sure no one got the wrong idea.

@JSE -- It sounds like joro was asking if it means there is a number field $K$ and we know that ranks are unbounded for elliptic curves over $K$... but in order to make a curve with larger rank in this way, you have to enlarge $K$. It does show that ranks are unbounded over some big field that contains all those guys, like the compositum of all quadratic extensions of the compositum of all quadratic extensions of $\mathbb{Q}$, which I would write as $\left(\mathbb{Q}^{(2)}\right)^{(2)}$.

I assume the polynomial has coefficients in $\mathbb{C}$. I think you can easily reduce this problem to the case where the polynomial has coefficients in a number field (like $\mathbb{Q}(i))$, so I added the nt tag.

By $p$-extension, do you possibly mean pro-$p$-extension (or am I confused by the meaning)?

@Will, Can you explain why you believe this?

Thanks, @Will. Here is how I am understanding the argument: If you look at the sum $S$ of $P - P^\sigma$ over all elements $\sigma$ of $\operatorname{Gal}(F(P)/F)$, then of course $S$ belongs to $E(F)$ because it is a sum of $p^n$-torsion points. On the other hand, if this Galois group has size $M$, then $S$ differs from $[M]P$ by an element of $E(F)$, so $[M]P \in E(F)$, and hence the order of $P$ in the quotient is at most $M$. It seems what I asked was too weak. The more difficult question would be to show that $[F(P):F]$ is only less than the size of $E[p^n]$ by a bounded amount.