This is not a full answer but easy to obtain. See the reference by @esg in the comments for the full answer.

I will give a very loose upperbound to this probability, which still tends to zero quite fast with increasing $N.$ You have $N^2$ balls thrown into $N$ bins where $N=2^n.$ So the probability that the lightest loaded bin has less than $N^\theta$ balls can be upper bounded (using the union bound in the first step)
$$
\mathbb{P}[Min<N^\theta]\leq
N \sum_{0\leq k < N^\theta} \mathbb{P}[X_1=k]=
N \sum_{0\leq k < N^\theta} \binom{N^2}{k} 2^{-N^2}
$$
where $X_1$ is the number of balls in the first bin. The binomial coefficients

$$\binom{N^2}{k}$$ are superincreasing in $k$ so upperbounding by the largest coefficient gives

$$
\mathbb{P}[Min<N^\theta]\leq N^{1+\theta} \binom{N^2}{N^\theta} 2^{-N^2}\sim
N^{1+\theta} 2^{-N^2(1-\mathbb{H}(N^{\theta-2}))}.
$$
by the entropy approximation to the binomial coefficient.
Using the crude bound $\mathbb{H}(p)< 2\sqrt{p(1-p)}$ we still get a bound that goes to zero exponentially fast.