What is the space of N=1 superVirasoro vertex superalgebras inside the c=n/2 free fermion vertex superalgebra? [Said differently, how many NeveuSchwartz vectors are there in n fermions?] Answers in terms of VOAs, CFTs, or nets all welcome, and any interpretation of "N=1 super" that makes you happy is okay.

$\begingroup$ Just to try to understand the question better, are you implicitly demanding that the free fermions be primaries wrt the resulting Virasoro vector? $\endgroup$ – José FigueroaO'Farrill Oct 27 '09 at 9:13

$\begingroup$ I'd like the Virasoro inside the superVirasoro to agree with the Virasoro of the free fermions (that is the nth tensor power of the Virasoro of the free fermion). The 'space' is then the space of extensions of the given Vir to an sVir. $\endgroup$ – cdouglas Oct 28 '09 at 4:12
I would be very surprised if this were not known, but I cannot find a suitable reference, so I did the calculation as the question seemed interesting to me. I believe that this is now the full answer to the question.
Let $V$ be an $n$dimensional real vector space with a symmetric inner product $\langle,\rangle$. Formulae are easier to write if we choose a basis $(e_i)$ for $V$ and let $$g_{ij} = \langle e_i,e_j\rangle.$$
Let $\psi_i$ denote the corresponding free fermion, with OPE $$\psi_i(z) \psi_j(w) = \frac{g_{ij} 1}{zw} + \mathrm{reg}$$
The standard Virasoro vector is then $$T = 1/2 g^{ij} \partial \psi_i \psi_j,$$ where the product is the normalordered product, which in my conventions associates to the left, so that $ABC = (A(BC))$. $T$ obeys the standard Virasoro OPE with $c = n/2$: $$T(z) T(w) = \frac{\frac{n}4 1}{(zw)^4} + 2 \frac{T(w)}{(zw)^2} + \frac{\partial T(w)}{zw} + \mathrm{reg}$$
The fields $\psi_i(z)$ are primary with weight $\frac12$ relative to $T$.
Then you are asking about the existence of a field $G$ which is primary of weight $\frac32$ relative to $T$ and whose OPE is $$G(z) G(w) = \frac{\frac{n}3 1}{(zw)^3} + \frac{2 T(w)}{zw}$$
The most general such $G$ takes the form $$G = \frac16 A^{ijk} \psi_i \psi_j \psi_k + B^i \partial \psi_i.$$
Calculating the relevant OPEs one sees that:
 for $G$ to be primary, $B^i=0$, and
 for $2T$ to appear in the firstorder pole of $GG$, $A$ must satisfy two conditions which I will now rephrase.
First of all, $A$ defines an alternating bilinear map $[,]: V \times V \to V$ by $$[e_i,e_j] = A_{ijk} g^{kl} e_l.$$
The existence of $G$ translates into the following conditions on this map:
 $\langle[x,y],z\rangle$ is totally skew
 $\langle[x,y],[z,w]\rangle$ is an algebraic curvature tensor
 and $$\mathrm{Tr}~ \mathrm{ad}_x \mathrm{ad}_y = 2 \langle x,y\rangle$$ where $\mathrm{ad}_x$ is the skewsymmetric endomorphism defined by $\mathrm{ad}_x(y) = [x,y].$
The algebraic curvature tensor condition means that the fourth rank tensor obeys the algebraic Bianchi identity: $$\langle[x,y],[z,w]\rangle + \mathrm{cyclic~in}~(x,y,z) = 0,$$ which using the invariance of the inner product under $\mathrm{ad}_x$ becomes $$\langle[[x,y],z],w\rangle + \mathrm{cyclic} = 0,$$ which in turn is equivalent to the Jacobi identity for the bracket. (Thanks for Paul de Medeiros for the observation.) Finally the last condition says that the Killing form, being twice the inner product, is nondegenerate, whence the Lie algebra is semisimple.
In summary, the solutions are in onetoone correspondence with real semisimple Lie algebras. If you further require the inner product to be positivedefinite, then these are the compact semisimple Lie algebras.
Interestingly, in this case one can embed an affine Lie algebra in such a way that the Virasoro vector coincides with the Sugawara construction. As I said above, I'm sure that this is standard, but cannot locate the reference right now.
EDIT This result is indeed known and can be found in: Goddard and Olive's KacMoody algebras, conformal symmetry and critical exponents, Nuclear Physics B, Volume 257, 1985, Pages 226252, in the very last section of the paper. (Thanks to Matthias Gaberdiel for the suggestion to look there.)
EDIT: I just noticed that the question asked "how many", whereas my answer just showed that as many as "semisimple Lie algebras". If, for the sake of simplicity, we take the inner product to be positivedefinite, then it is easy to write down a partition function and evaluating this gives a numerical answer to the question, depending on n. The first 100 values, starting at n=1, are the following:
0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 2, 3, 5, 3, 4, 8, 4, 5, 8, 7, 8, 11, 10, 11, 12, 13, 15, 19, 16, 21, 24, 21, 24, 32, 27, 34, 43, 37, 39, 53, 47, 54, 65, 65, 68, 79, 80, 90, 98, 102, 114, 129, 122, 138, 160, 157, 172, 207, 193, 211, 247, 244, 262, 306, 305, 329, 363, 378, 399, 448, 460, 505, 548, 554, 601, 675, 669, 739, 822, 826, 877, 990, 999, 1068, 1184, 1227, 1288, 1419, 1458, 1554, 1693, 1765
(A previous version of this edit had only taken into account the Aseries... apologies.)

1$\begingroup$ Thanks very much for this detailed answer! Much appreciated. (And sorry the thanks took so long in coming, my notification seems to be on the blink.) $\endgroup$ – cdouglas Dec 18 '09 at 8:39