Care to share the equation? I assume you respected all the limiting factors (group leaders will play away first, no teams from the same country can play against each other etc.)?
Darn, no, forgot about those. Hold on, let me recalculate.
Ok, I got 1 in 17088
Here's my logic.
Firstly, (for the curious) my original incorrect calculation of 1 in 2,027,025 came from simply the number of ways to pair up 16 teams when order is unimportant. The formula would be 16!/(8! * 2^8).
Now, we know that the 16 teams are actually separated into 2 groups of 8, and then 1 team from each group are drawn together. If you ignore country restrictions then the number of ways this can be done is 8! = 40320.
Now, how about the country restrictions. There are 6 possible matchups that would violate the country restrictions, although some are mutually exclusive. Let's label them:
A = Malaga vs Real Madrid
B = Malaga vs Valencia
C = Juventus vs Milan
D = Barcelona vs Real Madrid
E = Barcelona vs Valencia
F = Man U vs Arsenal
Now, we need to know how many of those 40320 ways do not contain any of the 6 above matchups.
The easiest way to do it, is to count how many of the 40320 ways do include at least one of those matchups, and then subtract that number from 40320.
That is, how many arrangements in the set A union B union C union D union E union F?
Using the inclusion/exclusion principle, the size of this union is equal to the sum of the sizes of all the singletons - the sum of the sizes of all the pairwise intersections + the sum of the sizes of all the triple intersections - the sum of the sizes of all the quad intersections + the sum of the sizes of all the 5-event intersections - the sum of the sizes of all the 6-event intersections.
Lets start at the end. Since there are two teams that appear in 2 different events, there is no way that all 6 events can ever happen simultaneously, likewise for any 5 of the 6.
There are 4 different combinations of 2 of the listed matches occurring simultaneously: ACEF, BCDF and each of those can occur in 4! ways since the other 4 matches are free to be drawn in any order.
There are 8 different combinations of 3 of the listed matches occurring simultaneously, and each of those can occur in 5! ways.
There are 11 different combinations of 2 of the listed matches occurring simultaneously, and each of those can occur in 6! ways.
Each of the 6 individual listed matches can occur in 7! ways.
Thus, the number of ways in which at least one of the 6 listed matches occurs is:
(6*7!)-(11*6!)+(8*5!)-(2*4!)+(0*5!)-(0*6!) = 23232
Thus, the number of valid ways (in which none of those 6 matches occur) is 40320-23232 = 17088.