Suppose that instead of the hexagonal pattern, we had the 127 cells laid out as three lines with 42, 43 and 42 cells.

So what?

There, the 41 cells without communication to the exterior would necessarily be hallways, otherwise, either the hallway would be two or more disconnected sections or some cell would not be near a hallway. Plus some other exterior cell would also need to be a hallway in order to connect it to the exterior, making up a minimum of 42 hallways cells.

But duh! It is not a three line pattern, it is an hexagonal pattern!

As long as you can fit it the hallways, it doesn't matter. If you can't fit, it would just make things worse.

Taking the three line pattern, we could make the entire middle line, with the exception of one of the cells in the very end as hallways. We could also exchange the entrance for some cell in the other two lines. But in any way, there would be a cell that we should call as the "end-of-hallway cell" and a non-hallway cell covered by it that we should call as "the last room". Surely, if the hallway bifurcates (as it does), there would be multiple cells eligible for being entitled as such, but we can choose any one of the multiple candidates and the result is the same.

Most commonly, each hallway cell connect to other two hallway cells. Some hallway cells are bifurcations having three other hallway cells as neighbors and some are dead-ends, having only one. However, every time we add a bifurcation, we also add a dead-end since cycles are inherent suboptimal. Hallway cells with four or more neighbors also being hallways can't be optimal since this would only increase the density of hallway cells, which is what we want to decrease.

So, with the exception of the end-of-hallway cell and the entrance, this implies that in the optimal case, the average number of neighbors that are hallway cells to other hallway cells neighbors is exactly two. Further, disregarding only "the last room", exactly a third of the cells must be hallways, otherwise some room would be inaccessible.

We have 127 cells. One of them is "the last room", so we have 126 other cells. Get a third of that and we have 42. Hence, this is best possible.

_{Thanks to Kruga for the sketch of the proof in a comment.}

Start with 0 hallways. Add the entrance in some edge. This will make 4 rooms accessible. If you start in a corner instead, this will give you 1 less accessible room and will only make your solution less optimal.

Starting add new hallways connected to an existing hallway. Every time you do that, the previous hallway will give access to 1 less room than it used to do. Every time you add a new hallway, at most 3 new rooms become accessible, but if that new hallway will not be a dead-end, one of those new rooms will be turned to the next hallway, so, with the exception of the entrance, each hallway section gives access to at most 2 rooms. Also, to get a new dead-end that gives access to 3 new rooms, you will also need to make a bifurcation that gives you 1 less room, so you still get 2 new reachable rooms per hallway in the best case.

This means that the best you can get is 2 new accessible rooms per hallway, and 1 more than that because the entrance started giving 4. So the number of cells is one more than the triple of the number of hallways. Or put in the other way around, the number of hallways is a third of the number of cells minus one. Hence, $(127 - 1) \div 3 = 42$. So this is the best solution possible.

_{Thanks to Kruga for the sketch of the proof in a comment and also for Daniel Wagner that also gave something similar in another comment.}

I got a solution with:

42 cells as hallways.

And I think that it is optimal, but I have no proof of that yet.

Here it is:

I got a solution with:

42 cells as hallways.

Here it is:

Proof of optimality

Suppose that instead of the hexagonal pattern, we had the 127 cells laid out as three lines with 42, 43 and 42 cells.

So what?

There, the 41 cells without communication to the exterior would necessarily be hallways, otherwise, either the hallway would be two or more disconnected sections or some cell would not be near a hallway. Plus some other exterior cell would also need to be a hallway in order to connect it to the exterior, making up a minimum of 42 hallways cells.

But duh! It is not a three line pattern, it is an hexagonal pattern!

As long as you can fit it the hallways, it doesn't matter. If you can't fit, it would just make things worse.

Taking the three line pattern, we could make the entire middle line, with the exception of one of the cells in the very end as hallways. We could also exchange the entrance for some cell in the other two lines. But in any way, there would be a cell that we should call as the "end-of-hallway cell" and a non-hallway cell covered by it that we should call as "the last room". Surely, if the hallway bifurcates (as it does), there would be multiple cells eligible for being entitled as such, but we can choose any one of the multiple candidates and the result is the same.

Most commonly, each hallway cell connect to other two hallway cells. Some hallway cells are bifurcations having three other hallway cells as neighbors and some are dead-ends, having only one. However, every time we add a bifurcation, we also add a dead-end since cycles are inherent suboptimal. Hallway cells with four or more neighbors also being hallways can't be optimal since this would only increase the density of hallway cells, which is what we want to decrease.

So, with the exception of the end-of-hallway cell and the entrance, this implies that in the optimal case, the average number of neighbors that are hallway cells to other hallway cells neighbors is exactly two. Further, disregarding only "the last room", exactly a third of the cells must be hallways, otherwise some room would be inaccessible.

We have 127 cells. One of them is "the last room", so we have 126 other cells. Get a third of that and we have 42. Hence, this is best possible.

_{Thanks to Kruga for the sketch of the proof in a comment.}