You're on the right track conceptually when you say it is "disjoint from the empty set." This isn't exactly how to phrase it, but the sentiment is there: it is disjoint from all the other cycles, because there are no other cycles.
This is how I would phrase it: a permutation $p$ is the product of disjoint cycles if, for some $n$, $p$ can be written as $p=\prod_{i=1}^nc_i$ where each $c_i$ is a cycle, and $c_i$ and $c_j$ are disjoint when $i\ne j$.
So, is $(123)$ a product of disjoint cycles? Let's test: is there an $n$ and a set of cycles $c_i$ so that $(123)=\prod_{i=1}^nc_i$? Sure: $n=1$ and $c_1=(123)$.
Let our permutation consists of $k$ cycles $(a^1_1\ldots a^1_{n_1})(a^2_1 \ldots a^2_{n_2})\ldots(a^k_1\ldots a^k_{n_k})$.
Then its Hamming distance to the cycle $a^1_1 \ldots a^1_{n_1} a^2_1 \ldots a^k_{n_k})$ is just $k$ (we need to change where $a^i_{n_i}$ go).
Now, let's count expected number of cycles of length $x$ in random permutation.
Probability of any given point to be in such cycle is $\frac{1}{n}$:
to have a cycle of length $x$ with point $a_1$ we need $a_1$ go to some point $a_2 \neq a_1$, probability of this is $\frac{n - 1}{n}$, then we need $a_2$ go to some point other then $a_1$ - probability of this, conditional on $a_1$ goes to $a_2$, is $\frac{n - 2}{n - 1}$, etc., and finally $a_x$ needs to go to $a_1$, probability of it conditionally on everything before is $\frac{1}{n - x + 1}$.
Multiplying, we get $\frac{n - 1}{n} \cdot \frac{n - 2}{n - 1} \cdot \ldots \cdot \frac{n - x + 1}{n - x + 2} \cdot \frac{1}{n - x + 1} = \frac{1}{n}$.
Number of cycles of length $x$ is equal to number of points in such cycles divided by $x$, expected number of points in such cycles is $1$, thus expected number of cycles of length $x$ is $\frac{1}{x}$.
Expected total number of cycles (of any length) is $\sum_{i=1}^n \frac{1}{i} = \log(n) + O(1)$.
It's probably should not be too hard to get much better estimation, but for your question it's enough to note that probability of having more than $4\log n$ cycles is less than $\frac{1}{2}$ (otherwise average number of cycles would be higher than $2\log(n)$).
So, random permutation with probability at least $\frac{1}{2}$ has at most $4\log n$ cycles, and thus is at distance of at most $\frac{4 \log n}{n}$ from a cycle.
And as $\frac{4 \log n}{n} < \frac{1}{4}$ for large enough $n$, we have that probability of random permutation to be $\epsilon$-close to set of cycles is at least $\frac{1}{2}$, which isn't $o(1)$.
Best Answer
We have from first principles that the number of permutations having maximum cycle length $k$ has EGF
$$\exp\left(\sum_{q=1}^k \frac{z^q}{q}\right) - \exp\left(\sum_{q=1}^{k-1} \frac{z^q}{q}\right).$$
This uses the combinatorial class
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\textsc{CYC}_{\le k}(\mathcal{Z}))$$
of permutations having cycle length at most $k.$ Therefore we get for the average maximum length
$$\frac{1}{n!} + [z^n] \sum_{k=2}^n k \left(\exp\left(\sum_{q=1}^k \frac{z^q}{q}\right) - \exp\left(\sum_{q=1}^{k-1} \frac{z^q}{q}\right)\right) \\ = \frac{1}{n!} + [z^n] \left( \sum_{k=2}^n k \exp\left(\sum_{q=1}^k \frac{z^q}{q}\right) - \sum_{k=1}^{n-1} (k+1) \exp\left(\sum_{q=1}^{k} \frac{z^q}{q}\right) \right).$$
Merging the two sums we get
$$\frac{1}{n!} + [z^n] \left(n \exp\left(\sum_{q=1}^n \frac{z^q}{q}\right) - \sum_{k=2}^{n-1} \exp\left(\sum_{q=1}^{k} \frac{z^q}{q}\right) - 2 \exp(z)\right).$$
This is $$\bbox[5px,border:2px solid #00A000]{ [z^n] \left(n \exp\left(\sum_{q=1}^n \frac{z^q}{q}\right) - \sum_{k=1}^{n-1} \exp\left(\sum_{q=1}^{k} \frac{z^q}{q}\right)\right).}$$
This gives the sequence $A_n$
$$1,3/2,{\frac{13}{6}},{\frac{67}{24}},{\frac{137}{40}}, {\frac{2911}{720}},{\frac{23563}{5040}},{\frac{23727}{4480}}, {\frac{2149927}{362880}},{\frac{23759791}{3628800}},\ldots$$
Computing the total sum of maximum cycle lengths of all $n!$ permutations we obtain $n! \times A_n$
$$1, 3, 13, 67, 411, 2911, 23563, 213543, 2149927, 23759791,\ldots$$
which points us to OEIS A028418 where these data are confirmed. The OEIS entry includes a recurrence which can be used for computational purposes (OEIS lists $450$ terms).
Owing to the coefficient extractor in front we can extend the first sum to infinity, getting
$$[z^n] \left(n \frac{1}{1-z} - \sum_{k=1}^{n-1} \exp\left(\sum_{q=1}^{k} \frac{z^q}{q}\right)\right)$$
or
$$\bbox[5px,border:2px solid #00A000]{ n - [z^n] \sum_{k=1}^{n-1} \exp\left(\sum_{q=1}^{k} \frac{z^q}{q}\right).}$$