Also a trained physicist. Don’t you need tension to stop the weight of the top ball pushing the 3 supporting balls outward?

My interpretation is that the writer meant close enough to all touch each other, in order to rule out non-triangular configurations (eg 3 balls in straight line with one balanced perfectly atop the center ball).

If you have high friction (the problem said "smooth" not "slippery"), then the only way the supporting balls can go anywhere is by rolling apart. But the ball sitting on top cannot simultaneously rotate in a manner compatible with all of the lower balls rolling away, so the lower balls would need to slip against the top ball if the top ball were to move downward.

In fact, even the signs are in favor of no motion -- the top ball (to the extent it moves at all) wants to fall straight down with no rotation, by symmetry. That motion would tend to rotate the top of each lower ball toward the center if you imagine the balls having high friction with each other or meshing like gears, which is the exact opposite of what they would need to do for anything to move. So you have a system where there's a factor (the tangential forces) trying to push the balls apart but another factor (friction plus rolling motion) trying to pull them together.

I suspect that any serious attempt to do the math here (factoring in all the rotational and tangential constraints) would discover that it's a statically overdetermined system with all the complications that such a system entails when asking questions like "how much tension is on this element?".

You need tension if the sphere-table contact is frictionless. But without friction the rope can’t stay on. If there is friction in the rope, there can’t be 0 tension in the rope before you put the top ball on because you need that tension to produce the rope/sphere friction.