Let the set of players be $N=\{1,\dots,n\}$.
According to my understand your description of the game, I take the following statements to be true:
- Each player chooses a number of players to integrate, not players with specific identities. As a result, the player making the choice may or may not be included in the coalition.
- Each player has to choose at least 1 player, if still available.
- Each player moves exactly once.
- The game is with perfect information.
Under the above interpretation, player $i$'s ($i\ge2$) action space would be $A_i=\{0,1,\dots,n+1-i\}$, as, in principle, $i$ could choose at most $n+1-i$ other players (when all players $j<i$ chose $1$) and at least $0$ (when all previous players have exhausted the player list). An exception applies to the first player, where the option $0$ is not feasible. Thus $A_1=\{1,\dots,n\}$.
Player 1's strategy space would be $S_1=\{1,\dots,n\}$. Player 2's would be $S_2=\{1,\dots,n-s_1\}$, since after $s_1$ players are picked by player 1, there are only $n-s_1$ players left to be picked by player 2. Carrying this argument forward, player $i$'s strategy space would be $S_i=\{1,\dots,n-s_1-\cdots-s_{i-1}\}$.
Generalizing from the above reasoning and taking into account boundary cases, let history at stage $i$ be $h^i=(s_0,s_1,\dots,s_{i-1})$, where we set $s_0=0$. (In general, though, stages should be indexed by a parameter different from the player index. But in your game, since each player moves exactly once, we can use the same index for stages as well as players to economize notation.) The history dependent strategy space for player $i$ would thus be
\begin{equation}
S_i(h^i)=
\begin{cases}
\{1,\dots,n-\sum_{j=0}^{i-1}s_j\} & \text{if } n-\sum_{j=0}^{i-1}s_j\ge 1 \\
\{0\} & \text{otherwise}.
\end{cases}
\end{equation}
