Finding all partitions of a set s.t. each block contains exactly one subset from a given set of subsets - how hopeless?
Here's an approach that I suspect will work very quickly.

I suggest you start by enumerating all possible values for an $S_i$, i.e., all possible 6-element subsets of $S$ that contain exactly one element of $N$ as a subset.

How many such sets do you expect to find?  By a crude back-of-the-envelope estimate, I estimate there will be only a handful, maybe 10 of them are so.  This estimate might be wrong, but this would be easy for you to test.

Let $\alpha$ be the number of possible such sets (i.e., $\alpha$ counts the number of ways to choose a 6-element subset of $S$ that contains exactly one element of $N$, i.e., there are $\alpha$ possible values for each $S_i$).  I am predicting you'll find $\alpha \approx 10$, or something similarly small.

If this prediction is accurate, this suggests a simple, naive algorithm to enumerate the partitions: namely, enumerate all ${\alpha \choose 5}$ ways to choose five valid possibilities for the $S_i$, and check whether those five sets forms a partition.  The running time of this algorithm is $O(\alpha^5)$.  If $\alpha$ is very small, say $\alpha=10$, this will be plenty efficient.

If it turns out that $\alpha$ is a bit larger, you can optimize this naive algorithm a bit.  Enumerate all ${\alpha \choose 2}$ ways to choose two out of these $\alpha$ 6-element sets, say $S_4,S_5$.  Filter out pairs that have an overlap (i.e., an element of $S$ in common).  Now store them in a hashtable, keyed by $S_4 \cup S_5$.  The hashtable will contain $O(\alpha^2)$ entries and can be constructed in $O(\alpha^2)$ time.  Next, enumerate all ${\alpha \choose 3}$ ways to choose $S_1,S_2,S_3$.  If $S_1,S_2,S_3$ are pairwise disjoint, look up $S \setminus (S_1 \cup S_2 \cup S_3)$ in the hashtable to see whether there is pair $S_4,S_5$ that can be combined with $S_1,S_2,S_3$ to form a full partition of $S$.  The running time of this optimized algorithm will be $O(\alpha^3)$.

And note that if $\alpha$ is indeed small, say $\alpha=10$ as I'm predicting, then it will be easy to enumerate all $\alpha$ possible values for an $S_i$.  You just enumerate all possible elements of $N$, then try extending each with three more elements of $S$, one by one, testing each one to make sure that the resulting set does not contain any other element of $N$ as a subset.  (In other words: select an element $T$ of $N$; select an element $x$ of $S \ T$, and check that $T \cup \{x\}$ does not contain any other element of $N$ as a subset; select an element $y$ of $S \setminus (T \cup \{x\})$, and check that $T \cup \{x,y\}$ does not contain any other element of $N$ as a subset; select an element $z$ of $S \setminus (T \cup \{x,y\})$, and check that $T \cup \{x,y,z\}$ does not contain any other element of $N$ as a subset.  Each step is pretty easy to do, and the overall running time should be very fast, assuming that $\alpha$ is indeed small as I expect.)

So, I suggest you start by running a little experiment to measure $\alpha$, then see if any of these algorithms is fast enough.  If it's not, edit your question to provide more details, including the value of $\alpha$.



Why do I estimate that $\alpha$ might be about 10?  Here's my crude back-of-the-envelope calculation.

First off, there are ${30 \choose 6} \approx 2^{19}$ ways to choose a 6-element subset of $S$, and ${30 \choose 3} \approx 2^{12}$ ways to choose a 3-element subset of $S$.  Therefore, $N$ contains about half of the possible 3-element subsets of $S$: in other words, if we pick a random 3-element subset of $S$, it'll be in $N$ with probability about 50% (roughly).

Also, for any 6-element subset $S_i$ of $S$, there are ${6 \choose 3} = 20$ ways to choose a 3-element subset of $S_i$, and each one has about a 50% chance of being an element of $N$.  We want there to be exactly one 3-element subset of $S_i$ that is in $N$.  This is like doing 20 coin flips, and we want to get exactly one heads (and the rest tails).  The probability of that is ${20 \choose 1} \times 2^{-20}$, i.e., $20/2^{20}$.  So, if we pick a 6-element subset of $S$, it will have probability about $20/2^{20}$ of being a valid possible value for some $S_i$.

Finally, there are $2^{19}$ possible 6-element subsets.  So, we expect that about $2^{19} \times 20/2^{20} = 10$ of them will be a valid possible value for $S_i$.  So, I'd expect $\alpha \approx 10$, using this crude argument.

This is a very crude estimate, because it assumes everything behaves randomly.  It is not rigorous, both because your set $N$ might not be random; and also because I made independence assumptions all over the place, even though things are not independent.  So, you will need to check empirically whether this estimate turns out to be accurate or not -- fortunately, that should be pretty easy to do.