The short answer, from a paper by Frank Ruskey, Carla D. Savage, and Stan Wagon is as follows:
... it is impossible to draw a Venn diagram with circles that will represent all the possible intersections of four (or more) sets. This is a simple consequence of the fact that circles can finitely intersect in at most two points and Euler’s relation F − E + V = 2 for the number of faces, edges, and vertices in a plane graph.
The same paper goes on in quite some detail about the process of creating Venn diagrams for higher values of n, especially for simple diagrams with rotational symmetry.
For a simple summary, the best answer I could find was on WikiAnswers:
Two circles intersect in at most two points, and each intersection creates one new region. (Going clockwise around the circle, the curve from each intersection to the next divides an existing region into two.)
Since the fourth circle intersects the first three in at most 6 places, it creates at most 6 new regions; that's 14 total, but you need 2^4 = 16 regions to represent all possible relationships between four sets.
But you can create a Venn diagram for four sets with four ellipses, because two ellipses can intersect in more than two points.
Both of these sources indicate that the critical property of a shape that would make it suitable or unsuitable for higher-order Venn diagrams is the number of possible intersections (and therefore, sub-regions) that can be made using two of the same shape.
To illustrate further, consider some of the complex shapes used for n=5, n=7 and n=11 (from Wolfram Mathworld):
The structure of these shapes is chosen such that they can intersect with each-other in as many different ways as required to produce the number of unique regions required for a given n.
See also: Are Venn Diagrams Limited to Three or Fewer Sets?
I prove this by assuming such a construction exists and deriving a contradiction.
Let's pick our two points of intersection for the construction and go from there. Since each has at least two lines passing through it, let us consider one of the lines ($L_1$) going through one point and two of the lines ($L_2, L_3$) going through the other. (If one of the lines happens to be passing through both points, we just pick the other three for the observation.)
Since neither of $L_2$ and $L_3$ intersect with $L_1$, they are both parallel to $L_1$ as well as each other. But since $L_2$ and $L_3$ pass through a common point, they must coincide, thus contradicting the hypothesis that there are 4 unique lines.
Best Answer
You know the fact that any circle can be characterised by any three points it passes through, i.e, If a circle passes through 3 points then it is unique. Therefore two distinct circles can intersect at no more than 2 points.
In case of ellipses, they are characterised by five points.