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This doesn't work as it stands on an edgeless board because the White King can hide around the shell in $O(1)$ moves rather than go after the Rook. But I think something similar should succeed, using a protected but pinned Black rook to substitute for the vertical edge.

EDIT See below for an explicit construction of mate in $N\omega$ with a fixed number of pieces on a ${\bf Z}^2$ board. TIDE

This doesn't work as it stands on an edgeless board because the White King can hide around the shell in $O(1)$ moves rather than go after the Rook. But I think something similar should succeed, using a protected but pinned Black rook to substitute for the vertical edge.

(I think the theoretical position Kc3,Qd1/Ka2,Rb2 is likewise a White win in $\log_2(n) + O(1)$ on an $n \times n$ board, and thus in $\omega$ on a quarter-infinite board, but the analysis is harder and it might be harder to adapt to an edgeless board.)

EDIT See my subsequent answer for a variant of this position with K+R vs. K+B+P on a quarter-infinite board thats mate in $2\omega$, and might be extended to $3\omega$, $4\omega$, etc. with more pawns. TIDE

(I think the theoretical position Kc3,Qd1/Ka2,Rb2 is likewise a White win in $\log_2(n) + O(1)$ on an $n \times n$ board, and thus in $\omega$ on a quarter-infinite board, but the analysis is harder and it might be harder to adapt to an edgeless board.)

On an infinite board even K+Q vs. K is not sufficient mating material against a lone King, while on a quarter-infinite board it is well known that K+R still suffice, with a mate of bounded length given the positions of the Kings (I think this is even in Winning Ways). Since mate in $\omega$ also requires Black to have a long-range piece, the minimum conceivable material is K+B vs. K+R. I claim that this is sufficient!

In the orthodox game K+B vs. K+R is usually an easy draw, but there are some known nontrivial wins. One standard example is Kb3,Rc2 / Kb1,Bc1. I claim that if we set this up on a quarter-infinite board with Black to move then White forces checkmate in $\omega$ moves.

On an infinite board even K+Q vs. K is not sufficient mating material against a lone King, while on a quarter-infinite board it is well known that K+R still suffice, with a mate of bounded length given the positions of the Kings (I think this is even in Winning Ways). Since mate in $\omega$ also requires Black to have a long-range piece, the minimum conceivable material is K+R vs. K+B. I claim that this is sufficient!

In the orthodox game K+R vs. K+B is usually an easy draw, but there are some known nontrivial wins. One standard example is Kb3,Rc2 / Kb1,Bc1. I claim that if we set this up on a quarter-infinite board with Black to move then White forces checkmate in $\omega$ moves.