The following published paper gives $20\times 2^{58}\approx 5.7646\times 10^{18}$:
Tomás Oliveira e Silva, "Empirical Verification of the 3x+1 and Related Conjectures." In "The Ultimate Challenge: The 3x+1 Problem," (edited by Jeffrey C. Lagarias), pp. 189-207, American Mathematical Society, 2010.
The author's website suggests ongoing work, but the sublink is down.
$$\frac{3e_0+1}{2^{\nu_2(3e_0+1)}}=e_1$$
can be rewritten as $$(3+\frac{1}{e_0})=2^{\nu_2(3e_0+1)}\frac{e_1}{e_0}$$
Now you have
$(3+\frac{1}{e_0})=2^{\nu_2(3e_0+1)}\frac{e_1}{e_0}$
$(3+\frac{1}{e_1})=2^{\nu_2(3e_1+1)}\frac{e_2}{e_1}$
...
$(3+\frac{1}{e_n})=2^{\nu_2(3e_n+1)}\frac{e_{n+1}}{e_n}$
You multiply every LHS/RHS to get
$(3+\frac{1}{e_0})(3+\frac{1}{e_1})...(3+\frac{1}{e_n})=\frac{e_{n+1}}{e_0}\prod_{k=0}^n2^{\nu_2(3e_k+1)}$
From here you get
$$(3+\frac{1}{e_{max}})^{n+1}\leq\frac{e_{n+1}}{e_0}\prod_{k=0}^n2^{\nu_2(3e_k+1)}\leq (3+\frac{1}{e_{min}})^{n+1}$$
But it means that in a cylce where $e_{n+1}=e_0$ you have
$\prod_{k=0}^n2^{\nu_2(3e_k+1)}\gt 3^{n+1}$ or $\begin{array}{|c|}\hline\sum\limits_{k=0}^n\nu_2(3e_k+1)>(n+1)\log_23\\\hline\end{array}$
Unless I messed something in translating to your notations, it does not match what you get.
Best Answer
Since nobody provided an answer to my question, I will answer myself.
As of August 2019, I am aware of ongoing BOINC project [1]. By personal correspondence with Eric Roosendaal I found that this ongoing BOINC project is meant to disprove the Collatz conjecture by trying to find a counter-example. The project started off in the middle of nowhere, at $2^{71}$ apparently, without specifying any arguments why this was chosen or why this would be a sensible point to use. It looks like they have reached roughly $2^{72.3}$ or so. No info is given as to whether all numbers up to that limit have indeed be checked.
As of August 2019, I am also aware of another ongoing project [2] by Eric Roosendaal. All numbers up to $2^{60} \approx 10^{18}$ have been checked for convergence.
In 2017, the yoyo@home project [3] [4] checked for convergence all numbers up to $10^{20} \approx 2^{66.4}$.
The paper by Tomás Oliveira e Silva [5] from 2010 claims that the author verified the conjecture up to $2^{62.3} \approx 5.76 \times 10^{18}$. Source: Tomás Oliveira e Silva, "Empirical Verification of the 3x+1 and Related Conjectures." In "The Ultimate Challenge: The 3x+1 Problem," (edited by Jeffrey C. Lagarias), pp. 189-207, American Mathematical Society, 2010.
The page [6] by Tomás Oliveira e Silva states that, in 2009, they verified the conjecture up to $2^{62.3}$.
Earlier, in 2008, Tomás Oliveira e Silva [6] tested all numbers below $19\times 2^{58}$.
Much earlier, in 1992, Leavens and Vermeulen verified the convergence for all numbers below $5.6 \times 10^{13} \approx 2^{45.67}$. Source: Leavens, G. T. and Vermeulen, M. "3x+1 Search Programs." Comput. Math. Appl. 24, 79-99, 1992.
By the way, the paper [7] from 2019 confirms to me that the largest integer being (consecutively) verified is about $2^{60}$, referring to above sources.
When I put it all together, I get the upper bound $2^{66.4}$.
UPDATE:
From September 2019 to May 2020, my project [8] managed to verify the Collatz conjecture for all numbers below $2^{68}$. So the current upper bound is $2^{68}$.