Is a CNF SAT problem NP hard when the total number (but not the width) of the 3-or-more-term clauses is bounded above by a constant? What about specifically when there's only one such clause?
It is worth noting that the problem becomes NP-hard when the restriction is relaxed slightly.
With a fixed number of clauses that are also of bounded size, the average number of literals in a clause is as close to 2 as one wants, by considering an instance with enough variables. As you point out, there is then a simple upper bound which is polynomial if the clause size is bounded.
In contrast, if the average number of literals per clause is at least $2 + \epsilon$ for some fixed (but arbitrarily small) $\epsilon > 0$, then the problem is NP-hard.
This can be shown by reducing 3SAT to this problem, by introducing new clauses with 2 literals that are trivially satisfiable. Suppose there are $m$ clauses in the 3SAT instance; to reduce the average clause size to $(2+\epsilon)$, it is enough to add $m(1 - \epsilon)/\epsilon$ new clauses with two literals. Since $\epsilon$ is fixed and positive, the new instance is of polynomial size.
This reduction also shows that even the version where the "large" clauses are restricted to 3 literals is NP-hard.
The remaining case is when the few large clauses are not of bounded size; each large clause seems to make the problem harder. See the SODA 2010 paper by Pǎtraşcu and Williams for the case of two clauses: they argue that if this can be done in sub-quadratic time then we would have better algorithms for SAT. There might be an extension of their argument to more clauses, which would provide evidence that your upper bound cannot be improved (modulo some form of the exponential time hypothesis).
Ok, I got it. The answer is no. This can be solved in poly-time. For each 3-or-more-term clause, select a literal and set it to be true. Then solve the remaining 2-sat problem. If any one provides a solution, then that is a solution to the overall problem. Since the number of 3-or-more-term clauses is fixed (say c), then if all such clauses have size <= m, then this runs in O(m^(c)*n). O(m^c) for going through each possible selection, times O(n) for solving the remaining 2-sat problem.