This will depend very much on the heat flow around the enclosure. There is a thing called the H factor which describes the heat transport properties between a surface and a fluid. The value of H varies with surface properties and flow.
So to do your calculation you need a few different assumptions. Let's simplify.
1) the air in the box is uniformly heated
2) air flows around five sides of the box the same way; no heat transfer takes place through the bottom
Now we can consider the box a "thermal resistance" (the walls) in series with a heat sink (the air flowing).
The thermal resistance (K/W) of the walls is given by
For your box, t = 1.6 mm, A = 3.4E5 mm$^2$, k = 17 K/W, so R = 2.8E-4 K/W : the walls themselves won't present much thermal resistance. So we need to look at the air around it.
The heat transfer coefficient is a strong function of air flow rate (this is why you have fans for cooling!). Let's make a simple assumption that there is some natural air flow, but no forced convection. The heat transfer coefficient will then be at the lower end of the scale, or about 0.5 W/m^2/K. (See for example this reference). With the "used" area of the box of 0.34 m$^2$ (12x16+(12+16+12+16)x6 in$^2$), you get about 1.5 K temperature rise inside the box for every Watt of power dissipated. A little fan around the outside could easily cut that by a factor 10...
So just to drive the point home explicitly: this calculation is conservative and VERY APPROXIMATE; the number calculated can be easily an order of magnitude off from the number you get when you build your box, because there are SO many factors that will affect this. Controlling the air flow explicitly (versus relying on "whatever air flow lies around") is really essential to get any kind of believable limits on your answer. This is reflected in the comments above...