I guess I was thinking that ECM would take
less power per mass on tungsten than on things like steel or copper, just because each atom is so much bigger. I'm not very clear on the chemistry, but suppose that we're electrolytically oxidizing each tungsten atom into an orthotungstate ion by withdrawing two electrons from it, with an electrolyte of something like chloric acid or sodium hypochlorite so we don't have to pay more electrons to free up some oxygens. Then at 100% Faraday efficiency we'd cut away 184 daltons of tungsten for every two electrons; contrast with iron where you still need to withdraw two electrons to get a stable water-soluble ferrous ion, but that only buys you 56 daltons of metal removal. Even copper, where the cuprous ion only costs one electron, you only dissolve 64 daltons.
There's also the issue of electrode potentials: sucking a coulomb off your anode at 0.5 volts takes only half a joule, but sucking off the same coulomb at 2 volts takes 2 joules. I don't understand standard electrode potentials well enough to predict what voltage you need for efficient tungsten ECM, but on first glance at the reference table, it doesn't look particularly bleak. The tungsten entries are in the middle of the table, near things like lead, tin, and iron, rather than way down on the upper east side with hoity-toity mercury and iridium, much less out in the ethereal districts where gold and ozone dwell.