For simplicity, I shall assign numbers to the double bonds so that it is easier to reference them. This numbering is chosen solely after their position in space and does not correspond to the number of carbons in a PIN or otherwise systematic name.
1 10 > 5 8 < 2 \ / 11 > 7 < 3 / \ 12 > 6 9 < 4 13
The terminal bonds, 1–4 and 10–13, are easy since they are all disubstituted: (1Z,2Z,3E,4E,10E,11Z,12Z,13E)
1Z 10E > 5 8 < 2Z \ / 11Z > 7 < 3E / \ 12Z > 6 9 < 4E 13E
Next up are the intermediate double bonds. If we simply follow the atoms, neither side will ever have priority, thus the somewhat less common rule that Z has priority over E takes effect. This allows us to label the double bonds (8Z) and (9Z).
1Z 10E > 5 8Z < 2Z \ / 11Z > 7 < 3E / \ 12Z > 6 9Z < 4E 13E
However, we cannot determine a stereodescriptor for bonds 5 and 6: both 1 and 2 have the same geometry (Z) as do both 3 and 4 (E). Thus, neither of these double bonds is asymmetric as they only feature three different substituents with the two identical ones being on the same side.
1Z 10E > 5Ø 8Z < 2Z \ / 11Z > 7 < 3E / \ 12Z > 6Ø 9Z < 4E 13E
Moving on to the final double bond 7, we face the same problem. The substituents 8 and 9 are identical, both feature an E double bond attache to E and Z. Thus again, we are unable to determine a stereodescriptor since the double bond is not asymmetric.
1Z 10E > 5Ø 8Z < 2Z \ / 11Z > 7Ø < 3E / \ 12Z > 6Ø 9Z < 4E 13E
Note that we can assign priorities to the left-hand side of double bond 7: the bottom half (3,4,6) has a lower priority than the top half (1,2,5) since the latter features Z double bonds where the former features E. But assigning a descriptor to 5, 6 and 7 is impossible due to their symmetric other side.