Quadrics containing many points in special position 
Suppose $n$ quadric hypersurfaces cut
  out $2^n$ distinct points
  $p_1,\ldots,p_{2^n}$ in
  $\mathbb{P}^n$. What is the maximal
  number of points $p_i$ a quadric can
  contain without containing all of
  them?

For n=2, there is of course a quadric going through any three points and avoiding the last point.
In $\mathbb{P}^3$, it is easy to exhibit a quadric going through 6 points (and avoiding the last two). This is however not possible with 7 points (as can be seen using a projection to $\mathbb{P}^2$).
 A: I do not know how much progress has been made on this, but what you ask is part of a conjecture appearing in a paper of Eisenbud, Green and Harris (see Cayley-Bacharach Theorems and Conjectures, Conjecture CB10).  The authors make the following conjecture.
Conjecture CB10. (Eisenbud, Green, Harris) Let $\Gamma$ be a complete intersection of $n$ quadrics in $\mathbb{P}^n$. If $X \subset \mathbb{P}^n$ is any hypersurface of degree $k$ containing a subscheme $\Gamma_0 \subset \Gamma$ of degree strictly greater than $2^n − 2^{n−k}$, then $X$ contains $\Gamma$.
If $X$ is a quadric, this says that if $X$ contains more than $3 \cdot 2^{n-2}$ of the points, then it must contain them all.
A: To complement the answers of Sasha and Damiano I would like to give one (completley non-generic)  example of a very specific configuration when a quadric contains exactly 
$3\cdot 2^{n-2}$ points.
Take in $\mathbb C^n$ the collection of $2^n$ points $(\pm 1,..., \pm 1)$. It is clear that this collection is the intersection of n quadrics. Indeed, we can obtain these points by taking intersection of $n$ degenerate quadrics $Q_i:=(x_i+1)(x_i-1)=0$, then we perturb a bit each quadric by adding a little multiple of $(\sum_i x_i^2-n)$. This way we get $n$ smooth quadrics that intersect transversally in the above $2^n$ points. 
Finally take one more singular quadric $(x_1-1)(x_2-1)=0$ it contains exactly $3\cdot 2^{n-2}$ points.
A: I think the answer is the stupid estimate $(n^2+n)/2$. On one hand, for any $(n^2 + n)/2$ points there is at least $n+1$ quadrics passing through it, just by counting the parameters. On the other hand, if you pick up $c_n := (n^2+n)/2 + 1$ points in generic position there are precisely $n$ quadrics through them. Now let $H_n$ be the irreducible component of the Hilbert scheme $(P^n)^{[c_n]}$ containing the subschemes consisting of $c_n$ distinct points. We have a rational map $H_n \to Gr(n,S^2V^*)$, $Z \mapsto H^0(P^n,I_Z(2))$ (here $V$ is the vector space of simension $n+1$ such that $P^n = P(V)$). It is clearly finite over generic point. Since $H_n$ is irreducible it follows that for generic $n$-tuple of quadrics any $c_n$ points in there intersection are generic on $H_n$, hence there are only $n$ quadrics passing through them.
