Suppose every clause contains all but c variables. I think the problem is in $P$, which I believe can be seem from your example with $c=0$. In your case, every clause "forbids" exactly one satisfying assignment. Suppose we have $n$ variables, then the formula is satifiable when there is at least one satisfying assignment that remains, and thus, when not all $2^n$ possible assignments are forbidden.
I believe we can generalize this idea: keep a list of all such "forbidden" assignments: for each clause, add the $2^c$ assignments it forbids to the list (the opposite of the clause, with all possible extensions to the missing variables). If in the end, the list contains all $2^n$ assignments, the formula is unsatisfiable. Else, it is satisfiable
Note that this is polynomial time, if there are $m$ clauses, the list contains at most $2^c \cdot m$ forbidden assignments.