I think you are asking for an objective measure of the effect of the rule. That needs to be answered with mathematical evidence, and little can be learned from particular character examples. Here I focus on the maths, and leave the comments for other people.
TL,DR;
The math involved in an exact answer is complicated, but the "feedback loop" (i.e. the recurrence) converges very rapidly to values very close what you get without inspiration just 0.25% above for rolling a 20 and below for rolling a 1. So, there will be a measurable but small consequence from this rule.
After many rolls, the probabilities of rolling n is approximately
$$ p(n) = (4.72 + 0.026 \times n ) \,\% $$
For example, if you start without advantage, the firts roll all numbers are equally probable (5%), the second roll the probability of rolling a 1 or a 20 are 4.76% and 5.23% respectively. After infinite rolls, where you use the inspiration immediately goes to 4.75 and 5.25% respectively. Very similar to the second roll.
Calculating the expected damage, can be done considering this probability. Just multiply the probability for each outcome with the corresponding damage, and add all the results.
Damage
The expected damage, or average damage, is the sum of all the possible damages weighted by their probability.
$$ \sum_{n=1}^{20} p(n) h(n) $$
where p(n) is the probability of rolling n, and h(n) is the damage rolling n, so zero if n<AC, and double if n=20.
For instance if the damage is (1D8 + 2), and a 10 hits, the average damage is 3.57500 without considering double damage on 20, 3.90000 with double damage on 20, and 3.99587 with inspiration rule and double damage on 20.
More simple, assuming a m and above hits, then the sum would be
$$ (\text{Damage}) \times \left( p(20) + \sum_{n = m}^{20} p(n) \right) $$
The factor on the right depend on m the roll that hits, depending on AC and modifiers, and is given in this table.
$$\begin{array}{cccc} \text{Roll that hits} & \text{Factor with Inspiration} & \text{Factor without Inspiration} & \text{Additional damage} \\\hline 2 & 1.005 & 1. & 0.48\% \\ 3 & 0.957 & 0.95 & 0.72\% \\ 4 & 0.909 & 0.9 & 0.97\% \\ 5 & 0.86 & 0.85 & 1.22\% \\ 6 & 0.812 & 0.8 & 1.47\% \\ 7 & 0.763 & 0.75 & 1.72\% \\ 8 & 0.714 & 0.7 & 1.96\% \\ 9 & 0.664 & 0.65 & 2.21\% \\ 10 & 0.615 & 0.6 & 2.46\% \\ 11 & 0.565 & 0.55 & 2.7\% \\ 12 & 0.515 & 0.5 & 2.95\% \\ 13 & 0.464 & 0.45 & 3.19\% \\ 14 & 0.414 & 0.4 & 3.44\% \\ 15 & 0.363 & 0.35 & 3.68\% \\ 16 & 0.312 & 0.3 & 3.92\% \\ 17 & 0.26 & 0.25 & 4.15\% \\ 18 & 0.209 & 0.2 & 4.38\% \\ 19 & 0.157 & 0.15 & 4.58\% \\ 20 & 0.105 & 0.1 & 4.75\% \\ \end{array}$$
Math analysis
Let's consider the D20 Discrete Uniform Distribution, any outcome is equally probable, so the probability is 5% for any number from 1 to 20.
$$ P_{D20}(n) = \frac{1}{20} $$.
| Plot 1: Discrete Uniform Distribution for a D20 |
 |
| This is the simplest case of rolling a D20, each outcome is equally probable. |
The "Advantage D20", that is taken by the maximum of two independent D20 rolls is given by
$$ P_{A D20}(n) = \frac{(2 \times n - 1)}{400} $$
This is no longer a fixed number, but a function of the number n. So for example the probability of getting a 1 with advantage is now 0.25% and rolling a 20 is 9.75%, compared with the fixed 5% without advantage.
| Plot 2: Discrete Distribution for a D20 with advantage |
 |
| By selecting the largest outcome from two dice the distribution now grows linearly from 0.25% for 1 to 9.75% for 20 |
Now consider you started without advantage and roll some unknown number, for the second roll (k=2), consider that there is an 5% (1/20) that the previous roll gave advantage and 95% (19/20) that it didn't. Now the "Inspiration D20" P(n,2) needs to consider this two cases, with and without advantage, weighted by the probability of each case.
$$ P_{I}(n,2) = 0.95 \times \frac{1}{20} + 0.05 \times \frac{(2 \, n - 1)}{400} = 0.047375 + 0.00025 \times n $$
| Plot 3: Inspiration Discrete Distribution |
 |
| Almost flat distribution, in mainly the Discrete Uniform Distribution and a bit of the Discrete Distribution for a D20 with advantage. |
Now we can generalize for the roll number k as a function of the outcome from the previous roll k-1,
$$ P_{I}(n,k) = \sum_{i=1}^{19} \left(P_{I}( i ,k-1) \times \frac{1}{20} \right)+ P_{I}( 20 ,k-1) \times \frac{(2 \, n - 1)}{400} $$
The recurrence here becomes complicated to do by hand, using Wolfram Mathematica
sol = First@ RSolve[
Flatten@Table[
{
p[1,n] == 1/20,
p[k,n] == Sum[p[k-1,i],{i,1,19}]/20+p[k-1,20](2 n -1)/400
}
, {n,1,20}
]
, Evaluate@Table[p[k,n],{n,1,20}]
, k
]
I'm omitting the long mathematical expression here, as it doesn't add much.
The solution depends on the roll number and on the initial condition, but stabilizes very fast to the limit when k is very big, which can be calculated taking the limit when k goes to infinity to obtain.
$$ p(n) = \left(\frac{600}{127} + \frac{10 n}{381}\right) \,\% $$
Now the table of values is as follows
$$
\begin{array}{ccc} \text{Roll outcome} & \text{Probability fraction} & \text{Probability percentage} \\
\hline
1 & \frac{181}{3810} & 4.751 \% \\
2 & \frac{182}{3810} & 4.777 \% \\
3 & \frac{183}{3810} & 4.803 \% \\
4 & \frac{184}{3810} & 4.829 \% \\
5 & \frac{185}{3810} & 4.856 \% \\
6 & \frac{186}{3810} & 4.882 \% \\
7 & \frac{187}{3810} & 4.908 \% \\
8 & \frac{188}{3810} & 4.934 \% \\
9 & \frac{189}{3810} & 4.961 \% \\
10 & \frac{190}{3810} & 4.987 \% \\
11 & \frac{191}{3810} & 5.013 \% \\
12 & \frac{192}{3810} & 5.039 \% \\
13 & \frac{193}{3810} & 5.066 \% \\
14 & \frac{194}{3810} & 5.092 \% \\
15 & \frac{195}{3810} & 5.118 \% \\
16 & \frac{196}{3810} & 5.144 \% \\
17 & \frac{197}{3810} & 5.171 \% \\
18 & \frac{198}{3810} & 5.197 \% \\
19 & \frac{199}{3810} & 5.223 \% \\
20 & \frac{200}{3810} & 5.249 \% \\
\end{array}
$$
which is not very different from the regular 5% of getting any number.
Now, going back to the fact that the probability depends on the roll number, we know it tends to the values shown before, but how is the probability the roll just after a roll we know started with advantage, or without. This is given in this plots.
If you start without inspiration
| Plot 4: Probability trend from first to 4rth roll starting without advantage |
 |
| Notice that all the probabilities start the same (5%) and spread quickly. |
If you start with advantage.
| Plot 5: Probability trend from first to 4rth roll starting with advantage |
 |
| Notice that the probabilities start speeded like the Discrete Distribution for a D20 with advantage but converge quickly to the smaller spread of the Inspiration discrete Distribution. |
We see here that regardless if we start with or without advantage by the second roll the probabilities are almost indistinguishable from the limit (settled) case.
