Yes – deciding whether a 0/1 matrix can be turned into a Hankel matrix by independently permuting its rows and its columns is an NP-complete problem (so, unless P = NP, no polynomial-time algorithm exists).

Why the problem is in NP
If two permutations σ (of the rows) and τ (of the columns) are given it takes Θ(n²) time to build the rearranged matrix σ M τ and to verify that every north-east / south-west (skew) diagonal is constant, so the problem is in NP.

From Hankel to Toeplitz and back
Reverse the order of the columns of a matrix (i.e. apply the fixed permutation ρ that maps column j to column n+1–j).
A matrix H is Hankel ⇔ H ρ is Toeplitz, and conversely a matrix T is Toeplitz ⇔ T ρ is Hankel.
Consequently

    (Toeplitz-reorderability)  ≤ₚ  (Hankel-reorderability).

If it is NP-hard to decide whether a matrix can be permuted into a Toeplitz form, the same immediately holds for the Hankel form.

Toeplitz-reorderability is NP-complete
The NP-completeness of the “reorderable Toeplitz matrix” problem was proved in

A. Malucelli, S. Nicoloso & M. Servilio,  
“Reorderable Matrices”, 
Theor. Comp. Sci. 263 (2001) 129–142.  

(Theorem 4 of the paper.)

They give a polynomial reduction from Exact-3-Cover (an NP-complete problem) to the following decision problem:

RTM   Input: an n×n 0/1 matrix M.  
Question: do there exist permutations σ, τ such that σ M τ is
Toeplitz (i.e. every NW–SE diagonal is constant)?

Because reversing the columns changes “Toeplitz” into “Hankel”, the same construction reduces Exact-3-Cover to the Hankel version:

HTM   Input: an n×n 0/1 matrix M.  
Question: do there exist permutations σ, τ such that σ M τ is Hankel?

Thus HTM is NP-hard. Together with the membership in NP shown above, it is NP-complete.

Consequences
• Your Stack-Overflow question asks for a polynomial-time algorithm; the result above explains why none has been found.
• Heuristics, exponential-time search, or algorithms that exploit special structure (e.g. restricted numbers of distinct rows, bounded bandwidth, etc.) are all that can be hoped for in the general case unless P = NP.

In short: “Hankelability’’ is NP-complete.