
> 
> 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 $\mathbb{P}^2$).



Quadrics containing many points in special position
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.


