The Alexander horned sphere is a closed embedding of $S^2$ into $S^3$ which is not flat because otherwise the Schoenflies Theorem would be true for it. However, not being flat is not the same as not being able to find a neighborhood deformation retract. I suspect the answer to the question is no, but I have no valid argument. To be precise, let $\iota:S^2\hookrightarrow S^3$ be the embedding whose image is the Alexander horned sphere. Is $\iota$ a cofibration?
We'll make use of the following.
If $X$ is an ANR and $j:A\subseteq X$ is a closed subspace, then $A$ is an ANR if and only if the inclusion $j$ is a cofibration.
I don't know a good reference for this, although I suspect it may be in Borsuk's book in some form or another. A reference I do have which will cover the case at hand is found in chapter 3 of Daverman's book Decompositions of Manifolds, where Theorem 6 states that
If $A$ is a closed subset of a metric space $X$, then $A$ has HEP in $X$ with respect to every ANR $Y$.
The point is that if $A$ is itself an ANR, then so is $A\times I$, and in particular also the mapping cyclinder $M_j=A\times I\cup X\times 0$. In this case we see that $j:A\subseteq X$ is a cofibration by noticing that it satisfies the HEP with respect to $M_j$ (i.e. we show that $M_j$ is a retract of $X\times I$).
The point of course is that $S^3$ is an ANR, as is any closed subspace $A\subseteq S^3$ which is homeomorphic to $S^2$.

2$\begingroup$ I learnt from this answer that a reference for your first statement (with a very similar proof as what you describe) is Proposition A.6.7 of Cellular structures in topology, by Fritsch and Piccinini. $\endgroup$ – Pierre PC Oct 15 at 19:36

$\begingroup$ I find this amazing. Thank you very much for the clear and nice answer! I presume on the other hand that at least the Alexander horned sphere cannot have a topological regular neighborhood as defined by Edwards here arxiv.org/pdf/0904.4665.pdf ? $\endgroup$ – daniel Oct 16 at 1:53