Let $a_k=\pi(10^k)-\pi(10^{k-1})$ be the number of $k$ digit primes. All of them (for $k \gt 1$) are from the $4\cdot 10^{k-1}$ integers in that range which are co-prime to $10.$ The heuristic is that, for $k$ not too small, there should be about $\frac{2^k}{4\cdot10^{k-1}}a_k$ primes in that range using only the digits $3,7$ because the $2^k$ numbers of that form have the advantage of being co-prime to $10$ but nothing else (i.e. there seems no reason to expect advantages relative to other primes, nor do there seem to be any that persist meaningfully for large $k$) . (But **see the end**) for more carefully thoughts)

Anyway, the predicted and actual counts up to 15 digits are

$ \begin{array}{ccccccccccccccc} 2.0& 2.1& 2.9& 4.2&
6.7& 11& 18.8& 32.6& 58& 104& 188& 343& 632& 1171
& 2182\\\ 2&2&4&3&5&10&16&30&53&87&185&365&591&1062
&2290\end {array} $

The ratios are $1.00, .95, 1.40, .71, .75, .91, .85, .92, .92, .84, .99, 1.06, .94, .91, 1.05.$ We can't prove that there are infinitely many primes of this form but we would be willing to bet that the ratio of actual to expected will never again be observed to lie outside the range $(0.8,1.2)$ or even $(0.9,1.1)$ (if we start a little further out.)

Here are statistics for the most frequent prime divisors of the $256$ $8$ digit members along with the (naively) predicted number , $\frac{256}{p}.$

$[3,85,85],[11,70,23],[7,37,37],[101,36,3],[37,24,7],[73,16,4],[137,16,2],$$[17,15,15],[13,15,20],[23,13,11],[19,13,13],[29,9,9],[31,8,8]$

Many are just about right. One can see why $11,101$ and the larger divisors of $111=3 \cdot 37$ $1001=3 \cdot 7 \cdot 13$ and $10001=73 \cdot 137$ enjoy an advantage (and $37$, $73$ have other advantages.) However the actual count of $30$ is not far off the predicted count of $32.6$, In part this is because many of the "exceptionally popular" divisors strike out the same number including $37 \cdot 101 \cdot 73 \cdot 137 =37373737.$ For some reason $13$ actually comes out rather low.

Furthermore, the distribution $\mod p$ (for $p \ne 2,5$) does become uniform as $k$ grows. This is also true for the distribution $\mod m$ where $m$ is any modulus co-prime to $10$ (in particular the various distributions $\mod p$ are mutually independent):

Consider the (connected) directed graph with vertices $\{0,1,\cdots,m-1\}$ and edges $[u,10u+3]$ $[u,10u+7]$ (computed $\mod m$). So each vertex has indegree and outdegree $2$. A $k$-step random walk starting at $0$ corresponds to generating a $k$ digit member of the set by coin flips and keeping track of the evolving congruence class $\mod m$. The distribution $\mod m$ can also be seen by giving each edge the weight $1/2$, taking the (doubly stochastic) weighted adjacency matrix $A$ and computing $A^k\mathbf{v}$ for $\mathbf{v}=[1,0,0,\cdots,0]^t.$ The dominant eigenvalue is $1$ with the constant eigenvector. As noted, the convergence to uniformity can be somewhat slow for some primes such as $p=11$ and $p=101$. One can see directly why this is. For $p=11$ this is also reflected by the fact that the digraph is a path with loops at $7$ and $9$. This graph for $p=101$ is a little more interesting. (A fuller justification could be given. $3,7$ are relatively prime to each other and to the base $10$, they are not congruent mod $p$ for any $p$ other than $2$. The case of digits $1,7$ has advantages and disadvantages concerning divisibility by $3$ according as $k$ is not or is a multiple of 3.)

There is nothing special about the pair $3,7$, the same reasoning applies to any of the pairs from $\{1,3,7,9\}$ with the obvious exception of $3,9$ (**and** an adjustment for $1,7$). And, with simple adjustments, any set of possible digits in any base (with greatest common divisor 1) should be expected to have, for not too small $k$, the number of primes not ruled out by "obvious" restrictions (including congruence classes of differences).

**after more cautious thought:** The fuller form of the heuristic (which Wadim and other people can no doubt express more accurately than I) is that there are about the number of primes that one would expect from random considerations after taking into account the "obvious". (So there should be an "obvious" reason for deviations, maybe in hindsight.) So integers $4k^2-1$ factor algebraically and are not prime. These $3,7$ numbers are automatically odd and coprime to $5$ but other than that there seems no reason to think that they have special properties relative to other primes (nor do they appear to aside from explainable transient ones.) However, I would have expected the (somewhat vague) analysis I gave below to apply as well to digits $1,7.$ Upon reflection (and clearly if had bothered to check) when $k=3j$ all $2^k$ $k$-numbers are multiples of $3$ so none are prime although for $k=3j+1$ and $k=3j+2$ one would expect $\frac{2^k}{2/3\cdot 4 \cdot 10^{k-1}}a_k$ primes because all $2^k$ members are sure to be co-prime to $3$ as well as $10$.