A misère game is one where a player with no moves wins. The object of the game is to be presented with a board position in which you cannot do anything.
If we imagine a two-player game where the players take alternate turns, then we can think of a game as being isomorphic to a set of games that we can choose to present to our opponent. This means we can represent any game fitting our criteria as a set.
Suppose the value to the player is $+1$ if they win and $-1$ if the opponent wins.
The game $\{\}$ has value $+1$ because the player cannot make any moves and therefore wins.
The game $\{\{\}\}$ has value $-1$ because the player must pick the game $\{\}$ and give it to the opponent, who then wins.
The game $\{\{\}, \{\{\}\}\}$ has value $+1$ because the player picks the game $\{\{\}\}$ (which has value -1), to give to the opponent.
Does every set in ZFC have a value as a misère game? I think the axiom of regularity, guarantees that there must be some kind of "trace" through any set which has a value in some sense, but I don't know whether it's possible to extend this to a value for the entire set.
This is mainly about set theory; the misère game business is a red herring. Note that the outcome of a misère game can be calculated recursively: It's $\mathcal N$, which you called "value $+1$", if there are no moves or there's a move to a $\mathcal P$-position. It's $\mathcal P$, which you called "value $-1$", otherwise. Therefore, all that matters is whether play in the set ends in finite time, so that the induction can get to the base case.
As you suspected, the axiom of regularity/foundation proves no infinite descending sequence of sets exists.
(As an aside, instead of relying on regularity/foundation, the graduate textbook "Combinatorial Game Theory" by Aaron N. Siegel just defines the class of long games inductively using ordinals, so that it doesn't matter if regularity/foundation holds.)