There is an explicit formula for the integral with respect to the Haar measure of any polynomial in the entries of a unitary and its conjugate, due to Collins and Śniady:
Benoît Collins and Piotr Śniady. Integration with Respect to the Haar Measure on Unitary, Orthogonal and Symplectic Group. Communications in Mathematical Physics 264: 773-795, 2006. [arXiv:math-ph/0402073]
I won't actually write down the formula in this answer, it can be found in the paper.
For high-degree polynomials you need to know some things about the representation theory of the symmetric group to evaluate it, but for low-degree polynomials it's easy enough to just look up the required values.
Let me rewrite the question using notation that matches with Collins and Sniady: we'll assume the rows and columns of the unitary $U$ we're integrating over are indexed by the integers $1,\ldots,d$, so $d = 2^n$, and we'll denote the $(i,j)$ entry of $U$ by $U_{i,j}$. The question asks for the value
$$
\int \bigl\vert \bigl(U^2\bigr)_{1,1} \bigr\vert^2 \mathrm{d}U
$$
where the integral is with respect to Haar measure for $d\times d$ unitary matrices.
We need to rewrite the integrand.
We have
$$
\bigl(U^2\bigr)_{1,1} = \sum_{i=1}^d U_{1,i} U_{i,1}
$$
and therefore
$$
\bigl\vert \bigl(U^2\bigr)_{1,1} \bigr\vert^2 = \sum_{i=1}^d\sum_{j=1}^d U_{1,i} U_{i,1} \overline{U_{1,j}} \overline{U_{j,1}}.
$$
From the formula of Collins and Sniady, we can conclude this:
$$
\int U_{1,i} U_{i,1} \overline{U_{1,j}} \overline{U_{j,1}} \, \mathrm{d}U
= \begin{cases}
\frac{2}{d(d+1)} & i = j = 1\\
\frac{1}{d^2-1} & i = j \not= 1\\
0 & i\not=j.
\end{cases}
$$
Therefore, the value we're looking for is
$$
\int \bigl\vert \bigl(U^2\bigr)_{1,1} \bigr\vert^2 \mathrm{d}U
= \frac{2}{d(d+1)} + (d-1) \frac{1}{d^2 - 1} = \frac{d+2}{d(d+1)}.
$$
Switching back to the notation in the original question gives
$$
\mathbb{E}\Bigl[ \bigl\vert \langle 0^n \vert U^2 \vert 0^n\rangle \bigr\vert^2\Bigr] = \frac{2^n + 2}{2^n(2^n+1)} = \frac{2^{n-1} + 1}{2^{n-1}(2^n+1)},
$$
which is ever-so-slightly larger than $1/2^n$.
So, unless I made a mistake in this calculation (which is a definite possibility — it should be checked carefully), the intuition that applying a unitary twice scrambles the probability more is wrong: it scrambles it slightly less.
It's kind of like vitamins: taking twice as much as you need isn't better for you, it's actually a little bit worse.
