Don gave a simple answer. Here's an excessively detailed answer that agrees:
314.30 covers handhole enclosures and tells us when applying 314.28(A) to treat the ends of the raceways as the "bottom" of the open bottom box.
As to 314.28(A), if you are splicing your conductors within the handhole enclosure, then the only regulated dimension is the depth of the box, the distance from the end of the conduit to the removeable cover, as per 314.28(A) Exception. The width and the length can be whatever you like that is sufficient for all the conduits and to provide enough space to work in and make up the splices.
If instead you are pulling conductors through the box without splicing, then your width and length will be determined by the requirement that "The distance between raceway entries enclosing the same conductor shall not be less than six times the metric designator (trade size) of the larger raceway." So each pair of 4" conduits containing the same conductors will need to be 24" apart (I assume that's a clear dimension, rather than center-to-center.)
For example, the obvious layout to use is a row of 4 conduits coming in on one side and a row of 4 conduits going out on the other side. If you want each 4" conduit to be say 1" away from the interior sides of the enclosure, and 1" away from others in the row, then each row would take up 4*4.5 + 5 * 1 = 23" of internal space, so you'd want one internal dimension to be at least 23". If each wire goes from a conduit straight across to the corresponding conduit in the other row, then the distance between rows would be need to be 24", and this internal dimension would need to be at least 2*1 + 2*4.5 + 24 = 35".
If you want to make it smaller, then one simple variation on the above would be to run the wires diagonally across the box instead of straight across. Each conduit in one row would have wires going into the conduit 2 up or down in the other row. That would give you 11" of offset between conduits containing the same conductors, so the space between rows would only need to be sqrt(242-112) = 21.33". So you could save 2.5" and get away with 23" x 32.5" interior dimensions. Of course, having the conductors cross over like that might be more trouble than it is worth.
Cheers, Wayne