By investigating the structure of the result I come to the nice general formula
$$
\left|V\right| = (-1)^n \left(1-\sum _{i=1}^n \left(\frac{2}{\beta _i}+\sum _{j=1}^{i-1}
\frac{\left(v_i-v_j\right){}^2}{v_i v_j \beta _i \beta _j}\right)\right)
\prod _{k=1}^n v_k \beta _k
$$
If you have Mathematica you can use the following code to check the result
n = 4;
V = Table[If[i == j, (2 - β[i]) v[i], v[i] + v[j]], {i, n}, {j, n}];
(-1)^n (1 - Sum[2/β[i] + Sum[(v[i] - v[j])^2/(
v[i] v[j] β[i] β[j]), {j, i - 1}],
{i, n}]) Product[v[k] β[k], {k, n}] == Det[V] // Expand
(* True *)
I believe that there is a proof for this formula but I don't find it yet.