Hypercube bisector: Does this alternating sum have a limit?

Question

The following sum is the volume of the hyperplane that bisects a $D$-dimensional cube, midway between opposite corners. Does it have a limit as $D\to\infty$? Numerically, the limit seems to be $\sqrt{6/\pi}$.
$$\left(\frac D2\right)^{D-1}\frac{D^{3/2}}{D!}\sum_{k=0}^{\lfloor D/2\rfloor}(-1)^k{D\choose k}\left(\frac{D-2k}D\right)^{D-1}$$

Given a hypercube with coordinates $\{0,1\}^n$, what is the $n-1$-volume of the set of points within it equidistant from $(0,0,...,0)$ and $(1,1,...,1)$ ?
This thought follows on from the question What is the maximum volume of $N$-D slice of an $M$-D hypercube?
For $n=2$, the diagonal has length $\sqrt2$. For $n=3$, the hexagon has area $3\sqrt3/4\approx 1.299$ and for $n=4$, I think it is an octahedron with volume $4/3$.
I suppose the bisector's vertices in general are those points whose coordinates sum to $n/2$ with $\lfloor n/2\rfloor$ coordinates equal to each of $0$ and $1$.
I did simulations for 2 to 10 dimensions.

1. The simplex with vertices at the origin and the standard basis,
   $(0,0,...,0),(1,0,...,0),(0,1,0,...,0),...,(0,0,...,0,1)$ has volume $1/D!$.
2. Regard the origin as the top and the other vertices as the base, the height is $1/\sqrt D$ and the volume is $base×height/D$, so the base is $D^{3/2}/D!$ This base is the positive part of the hyperplane $\sum x_i=1$.
3. To pick random points on that base, I chose $D-1$ uniformly distributed numbers between 0 and 1 and let $\{y_i\}$ be those numbers in sorted order. Let $y_0=0,y_D=1$ and $x_i=y_i-y_{i-1}$.
4. The frequency with which all coordinates $x_i$ were below $2/D$ let me estimate the volume of the hyperplane that bisects the hypercube.
5. Results of $10^7$ points for each dimension gave these estimates for the volume:

Dim Volume
2 1.4142
3 1.2988
4 1.3331
5 1.3385
6 1.3476
7 1.3520
8 1.3551
9 1.3585
10 1.3597

6. Take the $n-1$-simplex and scale it up by a factor $D/2$, its volume is now $(D/2)^{D-1}D^{3/2}/D!$. This is the plane that bisects the unit cube.
7. Trim from each vertex a smaller simplex of relative size $(D-2)/D$ to prevent each $x_i$ going above $2/D$.
8. When $D\ge5$ these corners overlap, so by Inclusion-Exclusion replace $D\choose2$ smaller simplexes of relative size $(D-4)/D$, and so on.
9. The result is $$\left(\frac D2\right)^{D-1}\frac{D^{3/2}}{D!}\sum_{k=0}^{\lfloor D/2\rfloor}(-1)^k{D\choose k}\left(\frac{D-2k}D\right)^{D-1}$$
   which gives $115\sqrt5/192$ and $11\sqrt6/20$ for the next two.