Finding palindromic primes is not very hard, but rather proving them and labeling them as Prime instead of PRP (Probable Prime) is the hard job. You can find the list of world's largest palindromic primes here
And the reason why proving most of them is time-consuming lies in odd digits in them. That is if you could express your number as $k*2^n +- 1$ then proving is much easier using several algorithms out there, but since palindromic primes usually have more odd digits than $0$ ones, then their finding procedure becomes too long.
As an example, the best palindromic number is of the following form (the smallest one is $101$):
$100000000000...00000000000000001$
It can be easily expressed as $5^n*2^n + 1$ and therefore easy to do primility test. Occurance of a non-zero digit at any position, cause the power of $2$ in the formula to be decreased.
So the best place for such a digits is near the center of number.
Again as an example where power of $2$ is near $n/2$
$100000000000...007700...00000000000000001$
Updated
And for generating the number you can build a simple formula like the following in Mathematica:
Palindromic[n_, digit_, position_] :=
(10^n - 1)/9 + (digit - 1) (10^position + 10^(n - position - 1))
In[1]:= Palindromic[20, 7, 9]
Out[1]= 11111111177111111111
Best Answer
The primes from $1000$ to $10^8$ with that property (and the corresponding root) are:
Code (Haskell):
Between $10^8$ and $10^9$, we have
$111$ more. So these beasts are rare, but there are enough.