In József Beck's book Combinatorial Games: Tic-Tac-Toe Theory, he states the following open problems ("unrestricted $5$-in a row" is Gomoku on an infinite board):
Open Problem 4.1. Is it true that unrestricted $5$-in-a-row is a first player win?
Open Problem 4.2. Is it true that unrestricted $n$-in-a-row is a draw for every $n\ge6$?
There are some relevant counterexamples in the more general setting of "positional games". The following information is from pp. 78-84 of József Beck's book.
A positional game is defined by a hypergraph $(V,E)$ where $V$ is the vertex set (or board) and $E$ is the set of (hyper-)edges (or winning sets), a collection of nonempty finite subsets of $E$. Two players take turns picking (previously unpicked) vertices; the game is won by the first player to succeed in picking all the vertices of a winning set.
The following example, attributed to Fred Galvin, illustrates what Beck calls "the Induced Extra Set Paradox". The board is $V=\{1,2,3,4,5,6,7\}$; the winning sets are $\{1,2,3\},\{1,3,4\},\{1,4,5\},\{1,2,5\},\{4,5,6\},\{4,5,7\},\{6,7\}$. This game is a draw, but it becomes a first-player win if the board is restricted to the subset $\{1,2,3,4\}$ with winning sets $\{1,2,3\},\{1,3,4\},\{1,4,5\},\{1,2,5\}$.
In Galvin's example the hypergraph is non-uniform, meaning that the winning sets are not all the same size. Here is a slightly larger example of the same phenomenon, attributed to Sujith Vijay, where the winning sets are all triples: the board is $V=\{1,2,3,4,5,6,7,8,9\}$, the winning sets are $\{1,2,3\},\{1,2,4\},\{1,2,5\},\{1,3,4\},\{1,5,6\},\{3,5,7\},\{2,4,8\},\{2,6,9\}$. This game is a draw, but the restriction to the board $\{1,2,3,4,5,6,7\}$ is a first-player win.
Beck states the following open problems concerning generalized Tic-Tac-Toe, analogous to your question 3:
Open Problem 5.2 Is it true that, if the $n^d$ Tic-Tac-Toe is a first player win, then the $n^D$ game, where $D\gt d$, is also a win?
Open Problem 5.3. Is it true that, if the $n^d$ game is a draw, then the $(n+1)^d$ game is also a draw?
Here's a start on improving the $81^{16}$ upper bound.
The total number of ways that gobblets can be placed on the board is no more than
$$\left({16\choose0}+2{16\choose1}+4{16\choose2}+8{16\choose3}+14{16\choose4}+20{16\choose5}+20{16\choose6}\right)^4\\=277{,}993^4\approx5.97\times10^{21}$$
That is, a state of the board is determined by specifying, for each size gobblet, how many of that size are on the board, where they are, and which of those positions are occupied by white gobblets (and which by black). For example, there are $16\choose4$ ways to pick squares on which to place $4$ small gobblets, and for each of those choices there are ${4\choose1}+{4\choose2}+{4\choose3}=4+6+4=14$ ways to pick which squares are occupied by white gobblets.
This is still only an upper bound, because some of these states are not realizable in actual play, for various reasons. In particular, it ignores the fact that there cannot be more small gobblets on the board than larger ones (because of the way that gobblets are allowed to enter into play). It also ignores the fact that the game may end before a given state could be reached. Nonetheless, $6\times10^{21}$ is several orders of magnitude smaller than $81^{16}\approx3.4\times10^{30}$, so at least it's a start.
Best Answer
Well, it's easy to say something about the case $c=2$. There is no strategy, since all you can do is alternate between the two colors. Since there is a path of length $2n-2$ (counting the edges) between any two points, no game can take more than $2n-2$ moves. It's easy to construct a game that takes $2n-2$ moves; just use a checkerboard coloring.
At the other extreme, it's easy to say something about the case $c=n^2$. If each node is a different color, then you can always extend by one node at each move, and you can never extend by more than one node, so an optimal strategy will take $n^2-1$ moves.
So now we just have to fill in all the cases in between....