Please have a look at this thread as it describes a schedule that you might consider, also it has a link to another thread that attempts to explain why it's not possible have the properties that you described - 10 opponents twice and the last opponent once.
If you must have 7 rounds of play in tables of 4, then I think the schedule below is the best you can do:
There are 66 possible pairs of players (12*11)/2=66, in the schedule:
9 pairs occur once
54 pairs occur twice
3 pairs occur three times.
7 4 11 5
1 2 8 6
10 3 9 12
4 7 1 9
10 8 2 5
6 3 12 11
1 12 5 6
4 3 9 2
11 7 8 10
6 5 11 9
3 8 1 7
12 2 10 4
9 2 6 7
3 10 1 5
12 4 11 8
12 5 8 9
11 3 1 2
10 7 4 6
1 11 10 9
5 7 2 12
6 3 4 8