jmc

The function $f(x)=\sqrt{1-\sqrt{2-\sqrt{3-x}}}$ is defined only when $$1-\sqrt{2-\sqrt{3-x}}\ge 0,$$or $$\sqrt{2-\sqrt{3-x}}\le 1.(\ast )$$Squaring both sides, we get $$2-\sqrt{3-x}\le 1.$$Then $$\sqrt{3-x}\ge 1.$$Squaring both sides, we get $$3-x\ge 1,$$so $x\le 2.$

Also, for $(\ast )$ to hold, we must also have $$2-\sqrt{3-x}\ge 0.$$Then $\sqrt{3-x}\le 2.$ Squaring both sides, we get $$3-x\le 4,$$so $x\ge -1.$

Hence, the domain of $f(x)$ is $\boxed{[-1,2]}.$