Assuming the forward volt drop across the LED is 3.4V (check this) then the resistor is taking 13 - 3*(3.4) = 2.8V
So that means each resistor is dissipating 65mW with a total of 28 resistors that gives 1.8W of pure heat. This is not including any heat you get from the LED. So yes it will get hot.
how come its showing lil over 13?
because it is a liner power supply and the regulation on those is not very good. You will find that when you draw the rated current from it it will drop down to 12V.
Yes if you can drop the voltage it will be better but the thing dropping the voltage will then generate that heat so the total heat dissipation will remain the same. It's just the board will not get as hot.