Running Pro Mini Continuous on 12v w/o Damaging Regulator
Summary:

Simple, effective, low cost solution - choose any 3:
A series resistor in the regulator input, of properly designed resistance and wattage will greatly reduce regulator temperature and increase overall reliability. Reason and design details given below. 

Detail:

Assuming a 5V processor dissiaption in the regulator is
(Vsupply - Vregulator_out) x Current
= (12-5)x 0.040 = 280 mW - usefully less than the 480 mW if you are dissipating all the power in the regulator.
For a 3V3 processor dissipation is (12-3.3)* 0.040 ~= 350 mW

"Too hot to touch" is a minimum of 55C and probably hotter - say 70C for now.
At 30C ambient delta T = (70-30) = 40 C.
Thermal resistanc = dT/Watts =40/0.280 = 140 C/W or 114 C/W at 3V3.

Both of those thermal resistances are very high. You do not describe what heatsinking is on the regulator but that suggests PCB copper only and not much of it. Even adding a very rudimentary heat sink should help immensely. If there is no obvious way to add a formal heat sink then adding a 'flag' of copper or brass by soldering should be doable.

Using Mark William's 220C/Watt figure (see datasheet for conditions).
Say 300 mW (see above).
Delta t = 220 C/W  x 0.3 W = 66 C rise.
Ambient is 30-32 so allow 35 for very minor safety factor.
66 C rise + 35 C ~= 101C - water boils.
And temperature inbox will rise mildly from energy dissipated so hotter again.
BUIT a series resistor inside the box or external  will make an immense difference. 

An easy and effective solution:

The regulator needs some "headroom" - some voltage above the Vout value to allow some dropacross the internal switch. Check the data sheet, but a modern LDO probably allows as little as 1V headroom. So, for 5V out you need 6V in. The extra 6v from 12V to 6V is not "needed" by the regulator. So, it can be dissipated in a low cost high reliability resistor.
Determine 


Vsupply_minimum - the absolute lowest value supply will assume.  
Vreg_in_minimum  I'll use 6V to start.  
I_load_maximum - you said 40 ma. This must be abs_max or if of very short duration must be able to be supported by capacitors. 


Then

Rmax = V/I = (V_supply_min-Vregin_nax) / Imax
= (12 - 6 )/.040 = 150 Ohm.
This is the absolute maximum resistor that meets the specification. A slightly smaller value will be safer. Say 120 R in this case.
Then
Pmax in R = I^2R = .04^2 x 120 = 192 mW.
A 500 mW resistor would be "safe enough" and a 1 watt one very safe.
Pregulator max = (6-5) x 0.040 = 40 mW, 14% of the 280 mW initially. 
A  W resistor with adequate cooling will be very safe. Safest is probably a purpose built air cooled wire wound. 

Note that the dissipated energy will raise internal ambient. A resistor is still an effective solution.

Place a suitable decoupling capacitor at the regulator input. Even if not normally used here it is needed due to the series input resistor raising the supply resistance. 

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Temperature rise inside box due to heat dissipation.

Summary: Using pessimistic assumptions for box thickness, available area and power dissipation an internal temperature rise of under 15 degrees C can be expected. As the regulator is now dissipating under 15% of the original dissipation due to the series resistor it will have a far longer operating lifetime. 

Notes:

Temperature rise (assuming no air transfer occurs) is limited by heat transfer through the plastic. This is controlled by the thermal conductivity of th e plastic, its thickness and its area.  Obviously:
 - Thicker plastic leads to higher temperature rise.  


More surface area leads to lower temperature rise. 
Higher thermal conductivity of the plastic leads to lower temperature rise 


So given Thermal conductivity = Ktc


Delta T = Watts x thickness / (Ktc x area)  


So Ktc = Watts x thickness /(Delta_Tx area)
So units of Ktc must be Watts/metre/degree_C
Which is indeed how Ktc is usually expressed. 

This table of Thermal properties of plastic materials lists Ktc for ABS as 0.17 W/m/K
This means that 170 mW will cause a 1 degree K (or C) rise when plastic ratio of thickness to area is 1.

Assume thickness of box = 5mm = much higher than is likely.
Box area for a box L x W x D = 2 x (LW + WD + DL)
Here  = L D W = 100mm x  50mm  25 mm or 0.1 x 0.05 x 0.025m
Area = 0.0175 m^2.
Use a smaller more pessimistic area of 0.01 m^2
Use a pessimistic heat dissipation of 0.5 Watts

Using the formula - Delta T = Watts x thickness / (Ktc x area)  from above 

Temperature rise = 0.5 x 0.05 / (0.17 x 0.01) ~= 15 degrees C (or K)
This is based on a thicker than likely box, smaller than actual area, and higher than expected energy dissipation. A 15C rise is bearable and actual is lower, and as the regulator now dissipates under 20% of original it will survive very well.