I suspect there isn't for some simple reason, but I could not find anything on it and if the opposite would hold in a more general sense, then that would solve this question.

$\begingroup$ Not really what you want, but worth mentioning anyway is that a trivial answer is the degenerate case of arithmetic progression where the difference between each element is zero. If the initial element is a palindrome, all elements are palindromes. $\endgroup$ – hvd May 10 '15 at 22:10
A fun question! But the answer is no.
Suppose for contradiction that there is an arithmetic progression $a,a+r,\dots,a+(k1)r$ of palindromes with $k$ sufficiently large.
Firstly, by working with every tenth element of the progression, we may assume without loss of generality that the spacing $r$ of the progression is a multiple of $10$. This fixes the last digit of every element of the progression to be constant, and hence the first digit of every element of the progression is constant also.
Now observe that three four numbers with the same first digit cannot be in arithmetic progression unless they all have the same number of digits (due to the lacunarity of the set of numbers with a fixed first digit). Hence all the elements in the progression have the same number $d$ of digits. As they have the same leading digit, this forces $r < 10^{d1}$.
In particular, we have $10^i \leq r < 10^{i+1}$ for some $i \leq d4$ (say) if $k$ is large enough. This implies that either the pair $(a,a+r)$ or $(a+r,a+2r)$ have the same first $di2$ digits (one can have carries in one of these pairs, but not both), hence the same last $di2$ digits, which implies that $r$ is a multiple of $10^{di2}$. Thus all elements of the progression $a+jr$ have the same last $di2$ digits, hence the same first $di2$ digits, but this contradicts the lower bound $r \geq 10^i$ for $j$ big enough (say $j \geq 10^4$).
One could work through this argument more carefully and get an explicit upper bound on how long a progression one can have (it looks like something like $10^8$ or so without being too efficient). Presumably this can be lowered somewhat.
[EDIT: there is presumably a fancier proof exploiting the very significant differences between the Archimedean metric and the 10adic metric, and in particular establishing that these metrics are not bilipschitz on arbitrarily long arithmetic progressions.]

1$\begingroup$ Fair enough; I've changed "three" to "four" to fix this. $\endgroup$ – Terry Tao May 9 '15 at 18:45