For two-value logic and two-place connectives, there are $2^2 = 4$ possible inputs $\{00, 01, 10, 11\}$. Each of these can be mapped to any of the two possible output bits. One operator which maps each of the possible inputs onto $0$ could be represented as $\{0, 0 ,0 , 0\}$. Similarly, an operator that maps $00$ and $11$ to $0$ and the other two inputs to one would be $\{0, 1, 1, 0\}$. This becomes a binary counting problem with the total number of operators equal to $2^4 = 16$.
For a two-value logic and three-place connective, there are $2^3 = 8$ possible inputs. Each of those can be mapped to any of the two possible outputs. $\{0, 0, 0, 0, 0, 0, 0 ,0 ,0 ,0\}$ would be the operator that maps to all zeros and $\{1, 0, 0, 0, 0, 0, 0 ,0 ,0 ,1\}$ would map $00$ and $11$ to $1$ but all the other inputs to zero. This too becomes a counting problem with the number of total operators equal to $2^8 = 256$. In general we see that an $N$ input operator has $2^N$ possible inputs and $2^{2^N}$ operators.
Using similar reasoning it seems like this reasoning could be generailized to the forumula $V^{V^N}$ for a $V$-valued logic and an $N$-place connector. Is this correct or have I made a mistake in my reasoning somewhere?