Solution:

We proceed by strong induction $n$, with the base case $n=1$ being vacuous. For the inductive step, consider the situation in which we have two numbers $x$ and $y$ remaining on the blackboard. Suppose the first one was written after $a-1$ operations, and the second one was written after $b-1$ operations, so that $(a-1)+(b-1)=n-2$. Then by the inductive hypothesis,
$$x \geq 2^{\frac{4 a^{2}-4}{3}}, \quad y \geq 2^{\frac{4 b^{2}-4}{3}}$$
Consequently, using convexity and the bound $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$, we have
$$x+y \geq 2 \cdot 2^{\frac{2\left(a^{2}+b^{2}\right)-4}{3}} \geq 2^{\frac{(a+b)^{2}-1}{3}}=2^{\frac{n^{2}-1}{3}} .$$
So $(x+y)^{4} \geq 2^{\frac{4 n^{2}-4}{3}}$ as needed.