I can inject these lattice paths into cyclically symmetric plane partitions in an $(n-1) \times (n-1) \times (n-1)$ box. CSPP's grow at $(27/16)^{n^2/2 + O(n)}$. In the first draft this answer, I suggested that this would also be right for $N(n)$, but now I don't believe this; I think the right value should be $A^{n^2+O(n)}$ for some smaller $A$. However, numerically, they still seem close.
I will follow the usual route to biject non-intersecting paths with rhombus tilings, which I wrote up in an earlier answer. Figure 1 in Brandt-Bruce-Brysiewicz-Krone-Robeva shows how to compute $N(5) = 24$ as noncrossing paths from $\{ (1,0), (3,0) \}$ to $\{ (0,1), (0,3) \}$. The image below shows the grid they use, and one such pair of paths on it, tilted and skewed for future convenience:
Slide those paths down one unit and cover the spread out paths with rhombi with vertical sides. The result is shown shaded in the figure below. For each vertex that was not on the paths, add a horizontal rhomus. We get a rhombus tiling:
The number of non-crossing paths $N(n)$ is the number of rhombus tilings of the resulting shape. Another way to think of this shape is that it is an $(n-1) \times (n-1)$ rhombus, with $\lceil \tfrac{n-1}{2} \rceil$ triangles deleted from each of its top two sides. I've shaded the deleted triangles below:
Now, given a rhombus tiling of this shape, we can take three copies of it and arrange them centrally around a central vertex. The deleted triangles can be paired off with each other. Thus, we get a bijection with $(n-1) \times (n-1) \times (n-1)$ cyclically symmetric plane partitions which use the rhombi shaded below and don't cross the solid lines. (I missed the second condition the first time.)
Thus, $N(n)$ is less than the number of CSPP's, and the number of CSPP's grows at rate $(27/16)^{n^2/2+O(n)}$. I used to think the growth rates would match, but now I don't.
First, numerical data. Here is a table (taken from Table 1 in BBBKR and from OEIS) comparing the two:
$$
\begin{array}{c|cccccccc}
n & 1&2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\
\hline
N(n) &1& 1 & 2 & 5 & 24 & 149 & 1744 & 26825 & 769408 \\
CSPP(n-1) &1 & 2 & 5 & 20 & 132 & 1452 & 26741 & 826540 & 42939620 \\
\end{array}.$$
Interestingly, $N(n)$ seems very close to $CSPP(n-2)$ for low values of $n$, but I think this is a coincidence.
Now, theoretical arguments. The shape of a typical plane partition in a large $n \times n \times n$ box was worked out by Cohn, Larsen and Propp. There is a continuous function on the hexagon which gives the limiting probability that a rhombus in a given position will be used, and each of the fixed rhombi in our problem is in the most likely orientation, so I expect those not to impose a major restriction. However, the solid lines which may not be crossed are major problems. In the corners of the hexagon, these lie in the "frozen" region where, with probability $1$, all rhombi have the same orientation. Thus, it is highly unlikely that the solid lines will not be crossed.
This rhombus tiling problem (with the deleted rhombi) fits into the general framework of papers like Kenyon and Okounkov "Planar dimers and Harnack curves" and Cohn, Kenyon and Propp "A variational principle for domino tilings", where we have a region of growing size with height function on the boundary approaching a limit. This would take a while to write up carefully, and I don't see how to get a closed answer from it so, at least for today, I'm going to stop.
Will Sawin asks whether I can prove a lower bound of the form $A^{n^2}$ in this way. The answer is yes. I found lattice paths easier to draw than rhombi this time.
Take the half of the paths closer to the corner at $(0,0)$ and make them hug the axes as closely as possible. Then connect the other half of the points to the points on the line segments $((n/2, n/4), (3/n4, n/2))$ and $((n/4, n/2), (n/2, 3n/4))$. (I may have off by one errors; see the image.) Then fill the shaded hexagon with paths however you please. The number of ways to fill the hexagon is the number of plane partitions in an $(n/4) \times (n/4) \times (n/4)$ box, which grows like $a^{n^2}$.
