Since the numerator and the denominator are (apparently) not independent, I'm not convinced that their expectation can be computed separately.
First of all, note that:
$$\DeclareMathOperator{Tr}{Tr}\begin{align}\Tr\left(\mathbb{I} \otimes \left|0^{n-1}\middle\rangle\middle\langle 0^{n-1}\right|U\rho U^\dagger\right)&=\Tr\left(\left|0^n\middle\rangle\middle\langle0^n\right|U\rho U^\dagger + \left|1\middle\rangle\middle|0^{n-1}\middle\rangle\middle\langle1|\middle\langle0^{n-1}\right|U\rho U^\dagger\right)\\&=\Tr\left(\left|0^n\middle\rangle\middle\langle0^n\right|U\rho U^\dagger\right) + \Tr\left(\left|1\middle\rangle\middle|0^{n-1}\middle\rangle\middle\langle1|\middle\langle0^{n-1}\right|U\rho U^\dagger\right)\end{align}$$
Furthermore, note that:
$$\Tr\left(\left|0^{n}\middle\rangle\middle\langle 0^{n}\right|U\rho U^\dagger\right)=a_{0,0}$$
and :
$$\Tr\left(\left|1\middle\rangle\middle|0^{n-1}\middle\rangle\middle\langle1|\middle\langle0^{n-1}\right|U\rho U^\dagger\right)=a_{2^{n-1},2^{n-1}}$$
when denoting:
$$U\rho U^\dagger=\begin{pmatrix}a_{0,0}&\cdots&a_{0, 2^{n}-1\\}\\\vdots & \ddots&\vdots\\a_{2^n-1,0}&\cdots&a_{2^n-1,2^n-1}\end{pmatrix}$$
Thus:
$$\frac{\Tr\left(\left|0^{n}\middle\rangle\middle\langle 0^{n}\right|U\rho U^\dagger\right)}{\Tr\left(\mathbb{I} \otimes \left|0^{n-1}\middle\rangle\middle\langle 0^{n-1}\right|U\rho U^\dagger\right)}=\frac{a_{0,0}}{a_{0,0}+a_{2^{n-1},2^{n-1}}}$$
For now, suppose $\rho$ is pure. Thus, it can be written as $V|0\rangle$ without loss of generality. Since $UV$ is Haar-random for a Haar-random $U$, without loss of generality, we can assume that $\rho=|0\rangle\langle0|$. $a_{0,0}$ and $a_{2^{n-1},2^{n-1}}$ are then the probabilities of measuring the states $|0\rangle$ and $\left|2^{n-1}\right\rangle$ respectively. We know that for a Haar-random state, its coefficients are drawn independtly from a complex Gaussian and then normalized. That is, we can write $a_{0,0}$ as $\frac{A_{0,0}^2+B_{0,0}^2}{M^2}$ and $a_{2^{n-1},2^{n-1}}$ as $\frac{A_{2^{n-1},2^{n-1}}^2+B_{2^{n-1},2^{n-1}}^2}{M^2}$, with $A_{i,i}$ and $B_{i,i}$ being independelty drawn from $\mathcal{N}(0;1)$. The expectation we want to compute is then:
$$\mathbb{E}\left[\frac{\frac{A_{0,0}^2+B_{0,0}^2}{M^2}}{\frac{A_{0,0}^2+B_{0,0}^2}{M^2}+\frac{A_{2^{n-1},2^{n-1}}^2+B_{2^{n-1},2^{n-1}}^2}{M^2}}\right]=\mathbb{E}\left[\frac{A_{0,0}^2+B_{0,0}^2}{A_{0,0}^2+B_{0,0}^2+A_{2^{n-1},2^{n-1}}^2+B_{2^{n-1},2^{n-1}}^2}\right]$$
The interesting point here is that, contrarily to $a_{0,0}$ and $a_{2^{n-1},2^{n-1}}$, $A_{0,0}^2+B_{0,0}^2$ and $A_{2^{n-1},2^{n-1}}^2+B_{2^{n-1},2^{n-1}}^2$ are i.i.d. As such, we have:
$$\mathbb{E}\left[\frac{A_{0,0}^2+B_{0,0}^2}{A_{0,0}^2+B_{0,0}^2+A_{2^{n-1},2^{n-1}}^2+B_{2^{n-1},2^{n-1}}^2}\right]=\frac12.$$
Indeed, for $X$ and $Y$ being i.i.d., the following holds:
$$\mathbb{E}\left[\frac{X}{X+Y}\right]=\mathbb{E}\left[\frac{Y}{X+Y}\right]$$
but one also has that:
$$\mathbb{E}\left[\frac{X}{X+Y}\right]+\mathbb{E}\left[\frac{Y}{X+Y}\right]=\mathbb{E}\left[\frac{X+Y}{X+Y}\right]=1$$
Thus, if $\rho$ is pure, this expectation equals $\frac12$. Suppose now that $\rho$ can be written as:
$$\rho=\sum_kp_k\rho_k$$
with $\rho_k$ being pure for every $k$ and the $p_k$ summing to $1$. The quantity one wishes to take the expectation of is:
$$\frac{\sum\limits_kp_ka_{0,0,k}}{\sum\limits_kp_ka_{0,0,k}+\sum\limits_kp_ka_{2^{n-1},2^{n-1},k}}$$
Using the same argument as before, these two random variables are i.i.d. Thus, the expectation value is still $\frac12$ in this case.
