A small test in MATLAB, for number of vertex $N = 100$, each component is a uniform random number in $[0,1]$:
N = 100;
p=rand(N,3);
tic;
T = delaunayTri(p(:,1),p(:,2),p(:,3));
t = T.Triangulation;
e1 = p(t(:,2),:)-p(t(:,1),:);
e2 = p(t(:,3),:)-p(t(:,1),:);
e3 = p(t(:,4),:)-p(t(:,1),:);
V = abs(dot(cross(e1,e2,2),e3,2))/6;
Vol = sum(V);
time_elapse = toc;
Result:
time_elapse =
0.014807
Vol =
0.67880219135839
I would say it is reasonably fast, if you wanna run it $10^6$ times, it only takes less than 3 hours. Here is what it is like:

Also I want to mention that in Professor O'Rourke's post he mentioned using determinant to compute tetrahedra's volumes, yet here I prefer using triple product. It is a naturally vectorized operation, more scalable than the built-in routine of determinant (or you can expand a $4\times 4$ determinant by hand :p). Here is another test for $N=10^5$, the result is
time_elapse =
3.244278
Vol =
0.998068316875714
with tetrahedra number $\approx 7\times 10^5$. Notice the total volume is pretty closed to $1$ for there are too many points clustered in $[0,1]^3$.