What amount of lift would be created in the fallacious "lift from equal transit time" theory?
The theoretical lift coefficient of a NACA 4412 at the critical angle of attack (about 16 degrees) is approximately 1.7.
\begin{equation}
c_\mathrm{l}=1.7
\end{equation}

For our purposes, that same airfoil has its upper "surface" about 1.03 times as long as its lower, which can be seen by using the distance formula on the coordinates or integrating the cambered NACA airfoil equation. Here is the distance-formula method applied to the Airfoil Tools numbers (thanks to ROIMaison for making me check this calculation):


On average, therefore, the fallacy would predict the air moving along the upper surface at about 1.03 times the velocity as along the lower in order to join up properly.

Making all the assumptions from the NASA link, Bernoulli's equation says
\begin{equation}
P_1+\frac{\rho V_1^2}{2}=P_2+\frac{\rho V_2^2}{2}
\end{equation}
Let's represent the lower surface with subscript 1 and the upper with subscript 2. The respective velocities are averages across the surfaces. We have
\begin{align}
P_1-P_2&amp;=\frac{\rho (1.03V_1)^2}{2}-\frac{\rho V_1^2}{2}\\
&amp;=\frac{609}{20000}\rho V_1^2\\
\end{align}
So the average difference in pressure coefficient (i.e., the lift coefficient) is given by
\begin{equation}
\mathrm{avg}(c_\mathrm{p_{ETT}})=\frac{609}{10000}\frac{V_1^2}{V^2}=c_\mathrm{l_{ETT}}
\end{equation}
where $V$ is the freestream velocity and ETT stands for equal transit time.

We have another expression for pressure coefficient that we can apply to get a final value. We end up with
\begin{equation}
c_\mathrm{l_{ETT}}=\frac{609}{10000}\frac{V_1^2}{V^2}=1-\frac{V_1^2}{V^2}
\end{equation}
or
\begin{equation}
\frac{V_1^2}{V^2}=\frac{10000}{10609}
\end{equation}
and therefore
\begin{equation}
c_\mathrm{l_{ETT}}=\frac{609}{10609}\approx.0574 \approx .034c_\mathrm{l}
\end{equation}
which is less than 5% of the actual value. However, it is important to keep in mind that the fallacy predicts no lift change with angle of attack (just one of the many reasons it is silly), and so the relationship between $c_\mathrm{l_{ETT}}$ and $c_\mathrm{l}$ varies with angle of attack for a given airfoil. For the 4412, they meet at just over a -5° AOA. For a symmetric airfoil, they meet at zero AOA. For any other airfoil, the calculation must be redone with the correct path lengths and actual lift coefficients, although magnitude of the result is not likely to change dramatically.