Clarification for argument in proof of search in N^1/3 queries with hidden variables/non-collapsing measurements
The short answer is that the state we wrote down is only an approximation in the large $n$ limit (which is all that is required to prove the upper bound). The state isn't normalized for finite $n$, but becomes normalized as $n \rightarrow \infty$. It looks like we either used a different approximation than what I will describe below, or made a typo - thanks for pointing this out! - but these details don't affect our proof.

The details:

After $t$ Grover iterations, the exact state of the system is
$\sin((2t+1)\theta) | x \rangle + \cos((2t+1)\theta) \Sigma_{y\neq x} \frac{1}{\sqrt{2^n-1}} | y\rangle$
Where $\theta$ is such that $\sin(\theta) = 2^{-n/2}$ (this exact solution is given in Nielsen &amp; Chuang Section 6.1.3).

Working with the exact expression isn't very informative about how this state behaves asymptotically. So most of the time, to analyze asymptotics, people use the approximation that for small $\theta$, we have that $\sin\theta \approx \theta$ and $\cos\theta\approx 1-\theta^2/2$. Using this approximation, you can show that if you set $t=2^{n/3 -1}$, then your resulting state is $\approx \frac{2^{n/3}+1}{2^{n/2}} |x\rangle +\left(1-\frac{1}{2}\left(\frac{2^{n/3}+1}{2^{n/2}}\right)^2\right) \Sigma_{y\neq x} \frac{1}{\sqrt{2^n-1}} | y\rangle$. Note we've lost the normalization because of our approximation - so if you plug in any $n\in\mathbb{N}$ as you did, you'll get a non-normalized state. However as $n\rightarrow\infty$ the state becomes normalized, because the approximation $\sin\theta\approx \theta$ becomes increasingly accurate.

From this, you can easily see that if you measure this state, you see $x$ with probability $\Theta(2^{-n/3})$. Hence if you could perform $\tilde{O}(2^{n/3})$ non-collapsing measurements, you'd see the marked item once with constant probability. This is all we're using about the state to prove our upper bound.