Yes, there is always an equilibrium.
This game is a stochastic game: the number of chips in the pot and the identity of the player whose turn it is to move serve as a state variable. Since the number of chips in the pot is bounded (between 0 and M), there are finitely many states and actions to each player. In fact, the game is a stochastic game with perfect information: the players move alternately, so there are no simultaneous moves. Such games have an equilibrium that do not involve randomization, that is, the choice of the number of chips is deterministic and depends only on the current state, see Thuijsman and Raghavan, Perfect Information Stochastic Games and Related Classes, International Journal of Game Theory, 1997, 26, 403-408.
Thuijsman and Raghavan (1997) also identify one such equilibrium. Denote by $\sigma_i$ a max-min strategy of player $i$: a deterministic strategy that yields the maximal payoff to player $i$ if all other players play against her (against player $i$) and try to lower her payoff. Denote by $\tau_{-i}$ a deterministic punishment strategy vector of all players except player $i$ who try to lower player $i$'s payoff. Then an equilibrium looks as follows: each player $i$ follows $\sigma_i$. They do so until the first time in which some player, denoted $i_*$, deviates from her $\sigma_{i_*}$. Afterwards, all players punish player $i_*$ -- they switch to $\tau_{-i_*}$.