Primes very much behave like random numbers once you erase the basic properties and you start looking at very large numbers. So your question is if probability of a large number having a property P is $\frac{1}{\ln(n)}$ and we erase one digit, what is the probability of a new number having the same property P.

The question here is quite obvious, and that is which digit you have chosen to delete. Probability that a number $m-k10^n$ has a property P again, directly correlates with $10^n$. So you are asking if $m$ is prime what is the probability of $m-k10^n$ being prime.

Again the average gap between primes is $\ln(n)$ meaning that if you take digits up to $k\log_{10}(\ln(n))$ meaning first few (with some constant k that we can find heuristically), even for a very large number, the rest will keep the probability of $\frac{1}{\ln(\frac{n}{10})}$ for being a prime, regardless which digit you have deleted.

For the first few digits the situation is very complicated and very likely not to be solved any time soon regarding the precise distribution of prime gaps on such a small scale.

So being a prime number, $p$, and deleting a little bit larger digit does not differ much from guessing any prime number that has one digit less. Therefore you can still expect that a new number will be prime with the probability of

$$\frac{1}{\ln(\frac{p}{10})}$$

which is quite large.

Now if you want it for each digit you come up with the probability of the order of

$$\left ( \frac{1}{\ln(\frac{p}{10})} \right)^{\log_{10}(p)-k\log_{10}(\ln(p))} $$

which is the expression of the form

$$\frac{1}{(d-1)^d}$$

where $d$ is the number of digits, explaining why it will be notoriously difficult to verify the existence of the prime with such property with a very large number of digits just by listing them.

The option here is, then, to search for numbers in a specific form so we eliminate quite some number of possible factors when we delete a digit.