While it is stated for a 3A total power output, it will not meet that current without over heating at any significant voltage. It should be able to supply about 7-10W continuously just fine though, which is easily more than enough for what you need.
I assume you're using a canon DSLR, from your description, and these all use lithium technology batteries, and im sure you know, when a battery is fully charged and when a battery is empty, the voltages are different, so the camera a should work with the input range of approximately 3.3v-4.2v, I would keep the input voltage (to the camera) towards the higher side, as cameras use power in a fairly non-continuous manner, so a sudden load may cause the input voltage (to the camera) to drop.
You should be able to find a very cheap replacement battery for your camera from ebay or the like, you can gut the battery and use it for the contacts, so you can nicely attach the external supply. I would suggest you attach the supply to the small PCB inside the battery pack where the battery terminals attach, and remove the battery, as the cameras have some circuitry that monitors battery health, and you may need this PCB to "fool" the camera into thinking it is OK.
I assume from the battery life of my DSLRs that they use a very small amount of current when idle, without the LCD on, somewhere in the order of 10ma, however I do not know their power consumption when using alternative firmwares, but I would assume it is no more than 500ma.
Assume that the buck converter is terribly inefficient (say, 50%) and that works out at something like 4W of required power for your 4v at 500ma, your 8.5v 1.7A "wall wart" is well capable of supplying this.
The variable resistor is a 10-turn trim pot, used to make fine adjustments, over the range of 3-33v that means there is about 1 full turn per 3 volts and since your desired voltage is at the far end of the scale, I am not surprised it took a large number of turns.