I know this is an old question, but here's a little information about the $\textsf{ZFC}$ case to anyone finding this question.

In my MO question, Juan kindly pointed out that Woodin has a result stating that it's consistent relative to a sharp for a Woodin limit of Woodins that every $\textsf{OD}(\mathbb R)$ game of length $\omega_1$ on the reals is determined. This result can be found in Neeman's book on long games, exercise 7F.15.

Beyond that, we can find a non-determined definable game of length $\omega_1+\omega$ on the reals (equivalently the integers). After $\omega_1$ many rounds, player I has played a sequence $X$ of reals. If $X$ contains a perfect subset then we can define a well-ordering of the reals via $X$ and thus a non-determined set of reals $A\subseteq \omega^\omega$. Player I is then supposed to spend his last $\omega$ moves to land in $A$, making the game non-determined. If $X$ did *not* contain a perfect subset then the last $\omega$ rounds is spent playing the perfect set game on $X$, which is non-determined as $X$ doesn't have the perfect set property. It seems like this game is $\Delta^2_2$, so that's at least a lower definability bound for inconsistency. When we get to length $\mathfrak c+\omega$ then we only have to consider the first case above, so that the game is (at most) $\Pi^1_2$ (for all strategies there's a move which makes the strategy non-winning).

I'm not sure what happens below these lower bounds, except that Borel determinacy at least ensures that all Borel games are determined, no matter the length (just play elements of ${^\alpha}\omega$ for $\omega$ many rounds instead of playing elements of $\omega$ for $\alpha\omega$ many rounds). Here's an attempt to illustrate the situation, where red indicates inconsistencies and blue indicates consistency relative to large cardinals. I'm not sure if more is known.

Thispointclass is consistently determined, andthatpointclass, that insuch and such technical senseis the "next one" is not) or rather something more informal? $\endgroup$