Well, to start with, a cube 10 feet on a side has a 1000 cubic feet. Reduce by 20% gives 8 feet on a side: 512 cubic feet. That's about half, I'd say.
Then, manufacturing cost is always divided into each car. 50,000 Model S had to pay for $5M development and mfg cost, which runs out to $10 per car. 350,000 Model 3 paying for another 3.5 million development (they've already done most of the work on the S) comes out to a dollar a car. I figure that same math could apply to all the rest, so half the profit per car (making $25K profit now, Model 3 would make $12K profit) would easily pay for needed profit.
Then we drop the price of building the battery by a third ($40K less a third is 27K) and make it smaller by a third from 300 mile to 200 mile (so our 27 K becomes < 20K), so the battery is about half price.
I don't see a problem. High level fit and finish, moderate size and range, for half price. Doable!
What you've described is an object that's 50% smaller, not 20%. Of course, the overall measurement reduction is going to vary... for instance I suspect it's going to be much easier to reduce the length and width of the car, but likely not the height much if at all.
But again, reducing materials cost by a given percentage doesn't reduce the production cost by the same. If it costs $1 to stamp, finish, and install a $1 piece of material, then it costs $1.80 to do the same for 20% less material... a 10% cost reduction.
Certainly R&D can be amortized over a larger production run, but typically for long-run high volume production, that's going to be an increasing smaller factor in overall price than manufacturing costs.