Assuming no tuner losses, and assuming uniform loss throughout the line (which isn't true in practice, yet can be a practical simplifying assumption, more below), feedline losses in decibels are:
$$ -10 \log \left(L {
1 - |\Gamma|^2 \over
1-L^2 \: |\Gamma|^2
}\right) $$
Where $L$ is the fraction of power transmitted by the feedline into a matched load:
$$ L := 10^{-\text{loss in decibels}/10} $$
and $\Gamma$ is the reflection coefficient:
$$ |\Gamma| := { \text{VSWR} -1 \over \text{VSWR} +1} $$
Derivation
Let's work an example and look at how it's derived.
Begin by calculating the loss of the feedline from the datasheet and the length. Let's say it's 0.46 dB per 100 feet, and 100 feet long, so feedline loss is 0.46 dB. Let's undecibel that and make it a ratio:
$$ L := 10^{-0.46/10} = 0.9 $$
Meaning, 90% of the power is transmitted, and the other 10% is absorbed by the feedline.
The math is easier if we express VSWR as a reflection coefficient $\Gamma$. It's a complex number but we don't care about the phase of the reflection (and we can't know it from just the VSWR anyway), so we'll calculate just the magnitude.
$$ |\Gamma| := { \text{VSWR} -1 \over \text{VSWR} +1} $$
Let's say for our example the VSWR is 2.35:1. That's a reflection coefficient of 0.4. This is a ratio of amplitude or voltage, so square it to obtain power: 16% of power is reflected, and the other 84% is accepted by the antenna.
And let's say forward power is 100 W. Furthermore, we'll assume the transmitter contains a tuner. When the tuner is adjusted so the transmitter sees a matched load, effective the tuner is inserting an impedance in the line which cancels the reflected wave, so the transmitter doesn't see any reflected power. Since the tuner (neglecting losses) has no resistive components to absorb any power and convert it to heat, this must mean all the reflected power that reaches the transmitter is reflected again back towards the antenna.
Now let's visualize what happens to that power as it reflects back and forth a few times.

Sum all the power that was accepted by the antenna (75.6W + 9.8W + 1.3W + 0.16W + ...) to explain how much of that 100W was accepted by the antenna, and the rest was lost to the feedline. Shown in the diagram, a total of 86.8W makes it to the antenna. That means losses were:
$$ 10 \log\left(86.8 \over 100 \right) = 0.61\:\mathrm{dB} $$
Through more iterations we can obtain a more precise answer, but you can see that after a few reflections the powers have become small enough to be negligible.
Loss in decibels is calculated from the ratio of input power to output power, so let's calculate the fraction of input power that's accepted by the antenna.
Whatever the incident power at the antenna, $|\Gamma|^2$ of it is reflected, and $1-|\Gamma|^2$ of it is accepted. If we want to calculate the power accepted on a specific reflection, we need only find the incident power, and multiply it by $1-|\Gamma|^2$.
To calculate the incident power, we can count the number of trips though the feedline that have happened so far, and the number of times a reflection off the antenna has occurred so far. Each trip through the feedline reduces the power by a factor of $L$, and each reflection off the antenna by a factor of $|\Gamma|^2$.
For the first reflection, it's just one trip through the feedline and zero prior reflections.
$$ L^1 \: (|\Gamma|^2)^0 \: (1 - |\Gamma|^2) \\
\tag{accepted, 1st reflection}$$
On the second reflection, it's three trips through the feedline, and one prior reflection.
$$ L^3 \: (|\Gamma|^2)^1 \: (1 - |\Gamma|^2) \\
\tag{accepted, 2nd reflection}$$
On the third reflection, it's five trips through the feedline, and two prior reflections.
$$ L^5 \: (|\Gamma|^2)^2 \: (1 - |\Gamma|^2) \\
\tag{accepted, 3rd reflection}$$
A pattern emerges: the power accepted at each reflection is the same at the previous reflection, multiplied by
$$ L^2 \: |\Gamma|^2 $$
These summed together is the total power accepted:
$$ \sum_{n=0}^{\infty} L(1 - |\Gamma|^2) (L^2 \: |\Gamma|^2)^n
$$
This is a geometric series, and can be simplified to
$$ \lim_{n \to \infty}
L(1 - |\Gamma|^2)
{
1-(L^2 \: |\Gamma|^2)^n \over
1-L^2 \: |\Gamma|^2
} $$
Because $L^2 \: |\Gamma|^2$ is less than 1, $(L^2 \: |\Gamma|^2)^n \to 0$ so the equation further simplifies to:
$$ L {
1 - |\Gamma|^2 \over
1-L^2 \: |\Gamma|^2
} $$
Limitations
A shortcoming of this model is real transmission lines don't have uniform loss.
Loss in a transmission line can be due to resistive loss, or dielectric loss. Resistive loss is proportional to the square of current, whereas dielectric loss is proportional to the square of voltage.
When the line is terminated in a mismatched load, standing waves form regions of high voltage alternating with regions of high current. The derivation above assumes losses occur uniformly in these regions, but in practice the resistive losses are more significant than the dielectric losses. Thus, the regions with high current have higher loss than the regions with high voltage.
For example, consider a short 50 ohm transmission line, say 1/10th of a wavelength long. This line could be terminated in either 5 ohms or 500 ohms, and in either case would experience an SWR of 10:1, and the method above would predict the same loss in either case. But in actuality, this line is so short the current and voltage anywhere along its length is more or less uniform.
In the case of the 5 ohm termination, current is going to be high, and the actual losses will be more than the equation predicts. With a 500 ohm termination, current will be low and actual losses will be less, possibly even less than the matched loss of the line. The higher the SWR, the higher the deviation from the equation.
If the line is electrically a multiple of half wavelengths long, there will be an equal number of high voltage and high current nodes. The effects cancel out, and actual losses are exactly as the equation predicts.
Also if the line is a large number of wavelengths long, there will be a large number of nodes so having one extra high voltage or high current node doesn't make a big difference overall.
The practical consequence is that the equation derived here is accurate, but not always precise. In a case such as a non-resonant multi-band antenna where the electrical length and load impedance are effectively random or unknown, this equation provides an accurate estimate. If the load impedance and electrical length are known, and especially if the electrical length is less than a couple wavelengths, it may be worthwhile to employ more precise methods.