Sorry, I think it is not true that $(X_2-X_1, X_3-X_1,...)$ are independent. Nevertheless, you can get an expression for the density of these variables: If $\Phi$ is a $C^1$-diffeomorphism (i.e. bijection and $\Phi$ and $\Phi^{-1}$ are both differentiable) then for any random variable $X$ with density $f_X$,
$$f_{\Phi(X)}(\tilde{x}) = f(\Phi^{-1}(\tilde{x})) |D\Phi^{-1}(\tilde{x})|$$
The map $(x_1,x_2,x_3) \mapsto (x_1, x_2 - x_1, x_3-x_1)$ is such a map. Hence,
$$f_{X_1,X_2-X_1,X_3-X_1}(x_1,\tilde{x}_2, \tilde{x}_3) = f_{X_1,X_2,X_3}(x_1,\tilde{x}_2+x_1, \tilde{x}_3+x_1) = f_{X_1}(x_1) f_{X_2}(\tilde{x}_2+x_1) f_{X_3}(\tilde{x}_3+x_1)$$
so
$$f_{X_2-X_1,X_3-X_1}(\tilde{x}_2, \tilde{x}_3) = \int_{\mathbb{R}} f_{X_1}(x_1) f_{X_2}(\tilde{x}_2+x_1) f_{X_3}(\tilde{x}_3+x_1) dx_1$$
from here you can also figure out the densities of $X_2-X_1$ and $X_3-X_1$ and it does not seem as if $f_{X_2-X_1,X_3-X_1} = f_{X_2-X_1} * f_{X_3-X_1}$...
However, now that we know $f_{X_2-X_1,X_3-X_1}$ we can write down an explicit expression for
$$P[X_2-X_1 < a, X_3-X_1 < b] = \int_{-\infty}^{a}\int_{-\infty}^{b} f_{X_2-X_1,X_3-X_1}(\tilde{x}_2, \tilde{x}_3) d\tilde{x}_2 d\tilde{x}_3$$
namely,
$$P[X_2-X_1 < a, X_3-X_1 < b] = \int_{-\infty}^{a}\int_{-\infty}^{b} \int_{\mathbb{R}} f_{X_1}(x_1) f_{X_2}(\tilde{x}_2+x_1) f_{X_3}(\tilde{x}_3+x_1) dx_1 d\tilde{x}_2 d\tilde{x}_3$$
Now you can reorganize integrals to
$$\int_{\mathbb{R}} f_{X_1}(x_1) \left(\int_{-\infty}^{a} f_{X_2}(\tilde{x}_2+x_1)d\tilde{x}_2 \right) \left(\int_{-\infty}^{b} f_{X_3}(\tilde{x}_3+x_1) d\tilde{x}_3 \right) dx_1 $$
and by substitution (for example for $i=2$) we get
$$\int_{-\infty}^{a} f_{X_2}(\tilde{x}_2+x_1)d\tilde{x}_2 = \int_{-\infty}^{a+x_1} f_{X_2}(x_2)dx_2 = \Phi(a+x_1)$$
so what you get is
$$\int_{\mathbb{R}} f_{X_1}(x_1) \Phi(a+x_1) \Phi(b+x_1) dx_1 $$
Doesn't look to good though... :-( Does that help?
