Extend bivariate to multivariate convolution formula? In reference to this post, the pdf for dependent random variables $X_1+X_2$ is given by:
$$f_{X_1+X_2}(z) = \int_{-\infty}^{\infty} f_{X_1,X_2}(x,z-x) \mathrm dx$$
How does this formula extend to the multivariate case $f_{X_1+...+X_n}(z)$?
 A: Let's follow Dilip Sarwate's instructions in the post you reference:

... the formula is obtained by writing $F_{X_1+X_2}(z)$ as a double integral of the joint density function over the specified region and then "differentiating under the integral sign."

Let the joint density of random variables $(X_1, X_2, \ldots, X_n)$ be given by $f$.  Then, by definition, the density of the sum $X=s(X)=X_1 + X_2 + \cdots + X_n$ is
$$F_X(x) = \Pr(X \le x) = {\int \cdots \int}_{s(\mathbf x) \le x} f(\mathbf x)d\mathbf x.$$
Assuming $F_X$ is differentiable at $x$, it has a density there given by 
$$f_X(x) = \frac{d}{dx}F_X(x).$$
To obtain a formula like the quoted one, apply Fubini's Theorem to express $F_X$ as a repeated integral,
$$F_X(x) = \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \int_{-\infty}^{x-(x_2+x_3+\cdots+x_n)}f(x_1,x_2,\ldots,x_n) dx_1 dx_2 \cdots dx_n,$$
differentiate under the integral, and apply the (first) Fundamental Theorem of Calculus to obtain
$$\eqalign{
f_X(x) &= \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \left(\frac{d}{dx}\int_{-\infty}^{x-(x_2+x_3+\cdots+x_n)}f(x_1,x_2,\ldots,x_n) dx_1\right) dx_2 \cdots dx_n \\
&= \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty f(x-(x_2+x_3+\cdots+x_n),x_2,\ldots,x_n) dx_2 \cdots dx_n.
}$$
Any of the variables can play the role of $x_1$, yielding $n$ formulas for the sum.

Alternatively, define $Y_{i} = X_1 + X_2 + \cdots + X_i$ and apply the two-variable formula (as written just above for the case $n=2$) recursively via the relation
$$Y_i = X_1+\cdots+X_{i-1}+X_i = (X_1+\cdots+X_{i-1})+X_i = Y_{i-1}+X_i$$
for $i=n, n-1, \ldots, 2$ to obtain
$$\eqalign{
f_X(x) &= \int_{-\infty}^\infty f_{Y_{n-1},X_n}(x-x_n, x_n) dx_n \\
&= \int_{-\infty}^\infty \int_{-\infty}^\infty f_{Y_{n-2},X_{n-1},X_n}(x-x_{n}-x_{n-1}, x_{n-1}, x_n) dx_{n-1} dx_n  \\
&\cdots \\
&= \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty f(x-x_n-x_{n-1}-\cdots-x_2,x_2,\ldots,x_n) dx_2 \cdots dx_n,
}
$$
giving the same result.
