Your numbers sound valid, but heat transfer is a lot more than a comparison of volume. In the case of forced convection, whether it be from a fan, or draft, the overall heat transfer is the area of the surface in question X the convection heat transfer coefficient X the temperature difference between the surface and the fluid (air in this case). The area inside the firebox is likely far greater than the channel that air is blown through outside the stove, but call it equal. In both cases, the 40 and the 165 cfm, the velocity isn't that great, or that different since the cross sectional area for the space the fan is blowing through is likely larger than the flue. Again, I'll call the area the same to be conservative, which makes the velocity of air 4 times higher for the fan. The flue conditions are known, and is about 2.5 mph at 40 cfm, so call the fan 10 mph at the same cross sectional area as the flue. Both of which are fairly slow, but I'll double the heat transfer coefficient for the fan to account for it. This essentially boils down to the variables outside the temperature difference being twice as high for the fan equation. All that's left now is to look at the surface temperature of the outside of the stove and the surface temperature of the inside of the stove. Since ambient air is the fluid in both cases, all that would have to be true is that the brick temp inside the stove is twice that of the outside surface. Seems easy enough to imagine.
In the end, I don't disagree that a fan moves a lot of air compared to an EPA stove flue, but we need to remember that the air being moved isn't the same. Perhaps the fan induced energy removal can exceed the flue losses when the air intake is throttled back, but I'd wager that in many situations that isn't true.