On values of $n\geq 1$ satisfying that for all primorial $N_k$ less than $n$ the difference $n-N_k$ is a prime number In this post I present a similar question that shows section A19 of [1], I was inspired in it to define my sequence and question. I am asking about it since I think that the problem that arises from my question harmonizes with this.
For integers $k\geq 1$ we denote the primorial of order $k$ as $N_k$, that is 
$$N_k=\prod_{r=1}^k p_r$$
where thus $p_r$ denotes the $r-th$ prime number. I add as reference that Wikipedia has the article Primorial.
In next paragraph $n\geq 1 $ denotes the integer variable.
Definition. Now we consider the values of $n$ making $n-N_k$ prime for all those primorials $N_k$ satisfying $2\leq N_k<n$.
I've calculated the first few terms from the definition. Our sequence starts as $$n=4,9,25,43,49,73\ldots...$$
To illustrate our sequence we add the example about why $25$ is in our sequence: the only primorials less than $n=25$ are $N_1=2$ and $N_2=6$, and we get that the integers $25-2$ and $25-6$ are both prime numbers.
I have no idea/intutition if this sequence has only a finite number of terms.

Question. I don't know if it exists only a finite number of positive integers $n's$ being $$n-N_k$$ a prime number for all primorial $2\leq N_k<n$. What work can be done to study if the sequence of $n's$ from this definition  should have a finite number of terms? Many thanks.

I am asking if it is possible to deduce a proposition, get some idea about it, or set some conjecture. 
I emphasize that I am asking about what work or statements/heuristics we can get for the Question, since it is a hard question (see the comments) I think that I should accept an available and useful answer.
References:
[1] Richard K. Guy, Unsolved Problems in Number Theory, Unsolved problems in Intuitive Mathematics Volume I, Second Edition, Springer-Verlag (1994).
 A: Instead of looking at primorials, one can look at so-called admissible sets. We say that a set of integers $S = \{h_1, \cdots, h_k\}$ is admissible if for all primes $p$, $S \pmod{p}$ is not a complete set of residues. Note that for all $p > |S|$, $S \pmod{p}$ is automatically not a full set of residues.
For example, the set $\{1,2\}$ is not admissible since modulo 2 the set represents a complete set of residues, whereas $\{0,2\}$ is admissible.
The conjecture then (originally attributed to Hardy-Littlewood) that for any admissible set $S$, there exists infinitely many integers $n$ such that each of $n + h_1, n + h_2, \cdots, n + h_k$ is prime. In fact the Hardy-Littlewood $k$-tuple conjectures predicts an asymptotic formula for the number of such $n \leq X$.
This conjecture, of course, is totally out of reach. What is known is due to the seminal work of Yitang Zhang, James Maynard and the two polymath projects: that for for $h_i \geq 0$ for $i = 1, \cdots, k$ and $h_i < h_j$ for $i < j$, whenever $h_k - h_1 \geq 246$ there exists infinitely many $n$ such that there exists two indices $i < j$ such that $n + h_i, n + h_j$ are both prime.
Maynard also proved the following which is out of reach using Zhang's approach: for each positive integer $m$ there exists a number $B(m)$ such that whenever $S$ is an admissible set of size $k$ exceeding $B(m)$, there exists infinitely many $n$ such that $n + h_i$ is prime for at least $m$ indices $i$. In fact, one can take $B(m) = O(e^{4m}m^3)$.
These results are known as bounded gaps between primes. They represent an extremely high achievement in prime number theory in recent times, perhaps the greatest results in a generation.
A: I will extend  Stanley Yao Xiao definition of admissible k-tuple.
Your conjecture is not true if we arrive to a $k-$tuple not admissible!
Suppose that the $m-$tuple $(0,h_1,\cdots,h_m)$ not admissible, then for $k>m$ and $a > 0$ if you put values $a, a+h_1,a+h_2,\cdots,a+h_{m-1}$ in the $k-$tuple, it will never be admissible!
Your conjecture will stop with the first not admissible $k-$tuple:
$n=9 : 3+(0, 4)$
$n=25 : 19+(0, 4)$
$n=43 : 13+(0, 24, 28)$
$n=49 : 19+(0, 24, 28)$

For $N_k < n \leq N_{k+1}$, you have a unique $k-$tuple $(0,h_1,\cdots,h_{k-1})$ and the next $(k+1)-$tuple will be of the form $(0,b,b+h_1,\cdots,b+h_{k-1})$
