TITLE: Quadrics containing many points in special position
QUESTION [5 upvotes]: 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 $\mathbb{P}^2$).

REPLY [4 votes]: 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.