Your "n-SAT" problem is just asking if all possible $2^n$ clauses are present. For this problem to be $NP$-complete, you would have to be able to reduce unsatisfiable instances of SAT to polynomial size unsatisfiable instances of this problem.
But an instance of "n-SAT" is only unsatisfiable when the number of variables is at most $\log N$, where $N$ is the input length (analogously, when the number of clauses is at least $2^n$, where $n$ is the number of variables). So your polynomial time reduction would have to decrease the number of variables from $N$ to logarithmic in $N$... good luck!
UPDATE: You ask why your set isn't NP-complete. Note that this is a loaded question. Your set is in $P$, as established earlier. If $P=NP$ then every nontrivial set $S$ that's in $P$ is actually $NP$-complete (nontrivial means: there is at least one string $x \in S$ and one string $y \notin S$). If $P=NP$ we have a polytime SAT algorithm. To reduce SAT to $S$, just call the SAT algorithm and output $x$ if your SAT algorithm says "yes", $y$ if it says "no". So to prove to you that your $S \in P$ cannot be $NP$-complete is tantamount to proving $P \neq NP$!