Hackerrank Gem Stones
Your solution looks pretty good. However, a few details can be changed to make your code more pythonic :

Unused variable

You can use _ to name a variable whose value is not used. In your case, it applies to the first loop variable x.

List comprehension

You can rewrite your initialisaion of rockList abusing list comprehension :

rockList = [set(raw_input())
    for _ in xrange(int(raw_input()))]


Simplifying the logic

At the moment, for each element in the first set, you look in how many other sets it appears to know if it is a gem. You can make things clearer by considering that by default it is a gem except if we don't find it in one of the other sets (and in that case, we can stop looping) :

for x in rockList[0]:
    is_gem = True
    for y in rockList[1:]:
        if x not in y:
            is_gem = False
            break
    if is_gem:
        gemElement += 1


Using Python good stuff

Using Python all builtin, we can write this all(x in y for y in rockList[1:]).

You can extract rockList[1:] to call it only once (note that the cute way to write this in Python 3 would be to use extended iterable unpacking). Your code becomes :

gemElement = 0
other_rocks = rockList[1:]
for x in rockList[0]:
    if all(x in y for y in other_rocks):
        gemElement += 1
print gemElement


Now, you can see that we can easily get the actual list of gems doing :

gems = (x for x in rockList[0]
    if all(x in y for y in other_rocks))


This is not an actual list but a generator expression, you'll need to call list(gems) if you want to see an actual list but we don't really care about the list, we just need the number of elements : len(list(gems)) or sum if you don't want to build an intermediate list in memory.

The code is now :

rockList = [set(raw_input())
    for _ in xrange(int(raw_input()))]
other_rocks = rockList[1:]
print sum(1 for x in rockList[0]
    if all(x in y for y in other_rocks))


One step back from the code

The problem we are trying to solve is linked to a common problem : computing the intersection of multiple sets. It is a generic enough problem so that we can google it and find this answer for instance.

Making the solution as concise as possible, one can write :

print len(set.intersection(*[
    set(raw_input())
    for _ in xrange(int(raw_input()))]))