As an addendum to @OlivierRozier's answer I'll show an estimate for an upper bound which is much sharper than that of Rhin - but only based on heuristics.
For $N$ (number of odd steps in my notation) and $S=\lceil N \cdot \log_2 3\rceil$ I conjecture that
$$ {1 \over C \cdot N \cdot \ln N} \lt S \cdot \ln 2 - N \ln 3 \tag 1
$$
which holds, with $C=10$, for all $ 2< N <10^{1\,000\,000} $ with a remarkable smooth fitting curve (tested with Pari/GP with according digits precision).
My derivations for $a_\min$, $a_\max$ and $\alpha=a_\text{mean}$ (denoting the according elements of a hypothetical cycle) are below; the result is that
$$ \alpha \qquad (=a_\text{mean}) \lt C \cdot N^2 \ln N / 3 \qquad \text{with } C=11 \tag {2a}$$
$$ a_\min \lt (1-(2/3)^N) \cdot C \cdot N \ln N \qquad \text{with } C=11 \tag {2b}$$
$$ a_\max \lt 1.5^{N-1} \cdot ( C \cdot N \ln N +1)\qquad \text{with } C=11
\tag {2c}$$
In the tayloring of simple forms for the key-values $a_\min,a_\max,\alpha$ I leave small errors out; that errors are captured by the increase of $C=10$ to $C=11$ in the formulae. I think, that process of calculation can be made more precise with more work invested...
Pictures:
In the following a plot of the empirical (rational) values for the key values as colored dots, the bounds giving continuous curves of related colors, and the Rhin-bound for the $\alpha$-value for $N=1 \ldots 100$ is inserted to compare the preciseness of the estimates:
To see wider range, but not be killed by oversizing there is another plot where I use the most critical cases $N$, but only that $N$ from the continued fractions of $\log_2 3$, so $N \in \{1,5,41,306,15601,79335,190537 , \ldots \} $
Here $N=190537$ is the most critical point for the upper bounds; to satisfy the bounds at this coordinate I needed to adapt $C=11$ to compensate the neglected error-terms in the definition of the key-values.
Note: the $a_\max$ values increase too much for a good display in the picture, so for the $a_\max$-values I document their logarithms, see the related y-scale on the rhs of the picture.
Definitions for the key-values $\alpha, a_\min, a_\max$
(I'm using my notation here, because it is much error-provoking to translate notations "on-the-fly", so to say, please bear with me)
For a cycle in the odd elements $a_k$ with $N$ steps $3x+1$ and $S$-steps $x/2$ we can state the basic equation
$$ 2^S = (3+{1\over a_1 })(3+{1\over a_2 })\cdots(3+{1\over a_N })
\tag {3a}$$
We can assume a value $\alpha = a_\text{mean}$ as a rough mean-value of the $a_k$ and determine its value by the collection of the equal parentheses as product
$$ 2^S = (3+{1\over \alpha })^N \tag {3b}
$$
From this the value of the rough mean can be determined
$$ 2^{S/N} = 3+{1\over \alpha } \tag {3c}
$$
$$ \alpha = {1 \over 2^{S/N} -3 } \tag {3d}
$$
This is the option to determine that value for one current selection of $N$ (and thus $S$). To have an upper bound functionally depending on $N$ we have to refer to the Rhin or to the Ellison bound, and develop differently:
$$ {2^S \over 3^N} = (1+{1\over 3 \alpha })^N \tag {3f}
$$
.... logarithmize ...
$$ S \cdot \ln 2 - N \cdot \ln 3 = N \cdot \log(1+{1\over 3 \alpha }) \tag {3g}
$$
apply Rhin-bound:
$$ {1 \over 457 N^{13.3}} \lt S \cdot \ln 2 - N \cdot \ln 3 = N \cdot \log(1+{1\over 3 \alpha }) \tag {3h}
$$
$$ {1 \over 457 N^{14.3}} \lt \log(1+{1\over 3 \alpha }) \\
{1 \over 457 N^{14.3}} \lt {1\over 3 \alpha } $$
arriving at
$$ \text{Rhin bound:} \qquad \alpha \lt 153 N^{14.3} \tag {3i}
$$
and because $\alpha$ is a rough mean for the $a_k$ we know that $a_\min$ must be smaller, and $a_\max$ must be larger than $ \alpha$.
Next, $a_\min$ for some given $N$ should be detected.
We assume, that -after we have already the idea of a rough mean $\alpha$ the $a_\min$ and $ a_\max$ should be identified by the "widest stretch" between the $a_k$, meaning $a_k={ 3a_{k-1}+1 \over 2}$ and thus the cycle should have the form of an "1-cycle" (in the Steiner/Simons-sense).
For such a cycle we have the formula for $a_\min $ ($=a_1$ by convention):
$$ a_\min = {3^N -2^N\over 2^S - 3^N}
$$
So, for a given pair $(N,S)$ we can give the value $a_1$ - which of course is mostly rational.
To give a upper bound as a function dependend on $N$ I propose again to try my "$N \ln N$"-bound .
Let's start with
$$ a_\min = {3^N -2^N\over 2^S - 3^N} \tag {4a} $$
Then adapting this for the "$N \ln N$"-bound conjecture:
$$
2^S-3^N = {3^N -2^N\over a_\min} \\
2^S/3^N = 1+{1 -2^N/3^N \over a_\min} $$
$$ S \ln2 - N \ln3 = \log( 1+{1 -2^N/3^N \over a_\min}) \tag {4b} $$
Getting now to the relation to $a_\min$:
$$
{ 1 \over 11 \cdot N \ln N} \lt S \ln2 - N \ln3 = \log( 1+{1 -2^N/3^N \over a_\min}) \\
{ 1 \over 11 \cdot N \ln N} \lt \log( 1+{1 -2^N/3^N \over a_\min}) \\
{ 1 \over 11 \cdot N \ln N} \lt {1 -2^N/3^N \over a_\min} $$
Finally $$
a_\min \lt (1 -2^N/3^N) \cdot 11 \cdot N \ln N \tag {4c} $$
and
$$ \text{ub } a_\min = (1 -(2/3)^N) \cdot 11 \cdot N \ln N \tag {4d}
$$
and to make the formula smoother, and for larger $N$ with vanishing error, we can simply say (conjecturally)
$$
\text{ub } a_\min = 11 \cdot N \ln N \\
a_\min \lt 11 \cdot N \ln N \\ \tag {4e}
$$
From that "1-cycle" structure we have also the formula for the $a_\max$ value. This is
$$ a_N = 1.5^{N-1} \cdot(a_1+1) -1 \tag {5a} $$
from where the upper bound follows by simple insertion:
$$
\text{ub } a_\max = 1.5^{N-1} \cdot 11 \cdot N \ln N \\
a_\max \lt 1.5^{N-1} \cdot 11 \cdot N \ln N \tag {5b}
$$
With a bit more handwork this could be made more precise; for me at the moment it seems to be enough to have a rough proposal for the upper bound of the $a_\max$ - which using the sharp "$N \ln N$"-bound is $(3/2)^N \cdot 7 \cdot N \ln N$ . This is larger than $(3/2)^N$ but only by the factor $N \ln N$ and thus obviously smaller than $3^N$ and if $N$ has not its few smallest values ($N \ge 22$ to be exact), is also smaller than $2^N$.
Well, repeated: this "$N \ln N$"-bound-estimate is only a conjecture, I don't have a proof so far.
