Change output voltage of a charger using the shunt resistor?
One quick and dirty way to drop the supply by 1 V is to put two high-current diodes in series with the supply + line.  Each diode will drop the voltage 0.65 V for a total of 1.3 V which will take the supply down to 58 V.  Make sure each diode has the current rating for whatever current you expect to draw from the supply.  And you need to provide adequate heatsinking for the diodes as they will generate 1.3 x 8 = 10.4 W of heat if you are drawing 8 A.

This said, it is a BAD idea to use an SLA charger to charge Li-ion, because of the unreliability of the charger.  You can put more diodes in series to get the voltage down to let's say 56 V and do away with the 4.2 V constant-voltage charging that Li-ion chargers provide to maximize storage, but still a cheap SLA charger can malfunction and feed let's say 65 V through the diodes and burst the cells.

It does not appear that the FET is controlling the output voltage because it does not have a hefty heatsink.  Most likely it is just an active breaker that protects the supply from shorts or excessive output current.  If this is a DC to DC step-down converter, you would need to adjust the primary side circuit to adjust the output voltage.

I think your best bet is to purchase a Li-ion charger, and also you must use a balance-protection board, if you don't have one already.  If you can't find a 14S board, you can adapt a 17S or 20S board, I believe.