Hi Markie,
A balanced schedule is not possible here, but the closest it is possible to come to meeting you criterion of 'everyone 2x' is to have 30 rounds. In the schedule below players 6, 9, 11 & 14 have one fewer game than all the others, but adding the match (6 11 vs 9 14) would give you everyone paired together exactly twice.
( 8 10 v 14 3) ( 5 12 v 4 13) (11 6 v 2 1)
( 1 11 v 7 10) (12 4 v 8 13) ( 3 9 v 5 2)
( 6 9 v 1 4) (11 5 v 8 7) (10 14 v 3 13)
( 7 6 v 12 2) ( 3 10 v 1 4) (14 8 v 9 13)
( 6 3 v 4 11) ( 2 9 v 10 12) (13 14 v 7 5)
( 2 11 v 7 9) ( 1 6 v 12 14) ( 8 4 v 10 5)
(11 3 v 13 2) ( 1 12 v 14 5) ( 9 8 v 7 6)
( 1 11 v 12 4) (14 10 v 3 5) ( 2 13 v 9 8)
( 7 11 v 10 6) ( 4 2 v 13 14) ( 5 1 v 8 3)
( 5 9 v 6 12) ( 8 3 v 7 14) (10 4 v 2 1)
( 4 6 v 7 13) ( 3 2 v 5 14) ( 8 11 v 9 12)
(12 1 v 2 7) ( 8 10 v 6 9) ( 3 4 v 13 11)
( 1 6 v 5 13) (10 9 v 12 2) (14 4 v 11 7)
( 7 12 v 13 10) ( 9 11 v 14 1) ( 6 3 v 5 8)
( 8 2 v 4 14) ( 7 9 v 1 13) ( 5 3 v 6 12)
(11 9 v 4 5) ( 3 2 v 13 7) ( 8 6 v 10 12)
( 2 9 v 4 5) (14 6 v 7 10) ( 3 1 v 11 8)
(12 13 v 8 1) ( 3 7 v 10 11) (14 2 v 5 6)
( 1 9 v 3 7) (10 4 v 12 8) (13 6 v 14 11)
( 9 1 v 13 3) (11 14 v 2 10) ( 4 7 v 5 12)
( 9 13 v 10 6) (14 12 v 11 5) ( 2 4 v 8 1)
(12 3 v 2 6) (14 8 v 10 1) ( 5 7 v 4 13)
( 6 8 v 2 7) (10 3 v 9 4) (12 11 v 1 5)
(13 11 v 5 8) ( 9 14 v 7 1) ( 6 2 v 3 4)
( 7 12 v 14 2) ( 3 1 v 6 13) (10 9 v 11 4)
( 5 10 v 7 1) ( 2 11 v 8 13) ( 3 14 v 9 12)
(10 13 v 12 11) ( 6 4 v 1 14) ( 9 5 v 7 8)
( 6 14 v 4 9) (12 8 v 3 11) ( 5 13 v 2 10)
( 2 5 v 1 10) (13 12 v 3 9) ( 4 8 v 7 14)
(10 11 v 6 5) ( 7 4 v 3 12) ( 1 13 v 8 2)
