Using Fourier analysis in the $d$ variable, we can get the optimal upper bound. As in Tao's argument, if there is a distribution of sequences for which each pair is $r$-probably increasing, there must be a single sequence $d_i$ :
such that:

$$\sum_{1\leq a <b \leq n} \sum_{1 \leq i,j \leq k, i\neq j} \frac{1}{j-i} 1_{d_i = a} 1_{d_j=b} >(1+ o(1)) (2r-1) k \log k$$

We may rewrite the left hand side as:

$$\sum_{1\leq a ,b \leq n} \sum_{1 \leq i,j \leq k, i\neq j} \frac{1}{j-i} 1_{d_i = a} 1_{d_j=b} 1_{a<b}$$

Now use the "carrying the 1" decomposition:

$$ 1_{a<b} = \frac{b}{n} - \frac{a}{n} + \frac{ a-b \operatorname{mod} n}{n}$$

The first two terms are pretty simple. The first term is:

$$\sum_{1 \leq i,j \leq k, i\neq j} \frac{1}{j-i} \frac{d_j}{n} = \sum_{1 \leq j \leq k} \frac{d_j}{n} \log \left( \frac{j}{k-j} \right) \approx k $$

and the second term is similar.

Now we've replaced $1_{a<b}$ with a convolution operator. So let's take the discrete Fourier transform:

$$\frac{1}{n}\sum_{0 \leq \xi \leq n-1} \sum_{1 \leq i,j \leq k, i\neq j} \frac{1}{j-i} e\left( \frac{d_j \xi}{n}\right) e\left( \frac{- d_i \xi}{n}\right) \left( \sum_{x=0}^{n-1} \frac{x}{n} e \left( \frac{x \xi}{n} \right)\right)$$

By the Hilbert transform inequality:

$$\left|\sum_{1 \leq i,j \leq k, i\neq j} \frac{1}{j-i} e\left( \frac{d_j \xi}{n}\right) e\left( \frac{- d_i \xi}{n}\right)\right| \leq \pi k$$

So the bound is

$$\frac{\pi k}{n}\sum_{0 \leq \xi \leq n-1}\left| \sum_{x=0}^{n-1} \frac{x}{n} e \left( \frac{x \xi}{n} \right)\right| \approx \frac{\pi k}{n} \sum_{0 \leq \xi \leq n-1} \frac{n}{2 \pi \min(\xi,n-\xi)} \approx k \log n$$

This gives:

$$k \log n > (1+o(1)) (2r-1) k \log k$$

Hence:

$$n > k^{ 2r-1 +o(1) }$$

Note that the base notation trick in the original post allows us to remove the $o(1)$ in the exponent bound by amplification.

Here is the matching lower bound:

Suppose that I have two sequences of $k$ numbers, one of which is a sequence of numbers from $1$ to $n_1$ that is increasing with probability $r_1$, and one which is a sequence from $1$ to $n_2$ that is increasing with proobability $r_2$. For any fraction $a/b$, I can construct a sequence of $k^b$ numbers from $1$ to $n_1^a n_2^{b-a}$ that is increasing with probability $\frac{a}{b} r_1 + \frac{b-a}{b}r_2$ as follows:

Write a number from $1$ to $k^b$ in base $k$ notation as $b$ numbers from $1$ to $k$. Choose randomly $a$ of the numbers and apply the first sequence, getting a number from $1$ to $n_1$. For the rest, apply the second sequence, getting a number from $1$ to $n_2$. Encode this sequence of numbers lexicographically as a single number from $1$ to $n_1^a n_2^{b-a}$ (generalizing base notation).

For any two numbers $i$ and $j$, consider the first place where they are not equal. In that place, $j$ must be larger than $i$. $d_j$ is equal to $d_i$ in all previous places. Hence as long as $d_j> d_i$ in this place, $d_j>d_i$. The probability that it is greater in this place is at least $r_1$ if this number was chosen and $r_2$ if it wasn't, so the total probability is at least $\frac{a}{b} r_1 + \frac{b-a}{b}r_2$.

This shows that we can take $k=n^{1/f(r)}$ for a convex function $f$. In particular, because we can take $f(1)=1$ (totally deterministic sequence) and $f(1/2-\epsilon)=0$ (totally random), we know $f(r) \leq 2r-1+\epsilon$, so we can get $k$ at least $n^{1/(2r-1)-\epsilon}$

An explanation of why $1/(j-i)$ is in fact the right weight function. Basically, this lower bounding method spends an equal amount of work improving the probability that $d_j>d_i$ for $j-i$ at any given scale. So the weight function should make every scale equally valuable, which $1/(j-i)$ does.