$\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}$The answer is $\tfrac{\pi^n}{n!}$. It is certainly a surprise to have the answer come out so simple!

Let $\phi : \CC^n \to \CC^n$ be the map which takes $(z_1, z_2, \ldots, z_n)$ to the elementary symmetric functions $(e_1, e_2, \ldots, e_n)$ where $e_k = \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq n} z_{i_1} z_{i_2} \cdots z_{i_n}$. Let $D$ be the unit disc in $\CC$. So you want to compute the volume of $\phi(D^n)$, also known as $\int_{\phi(D^n)} \mathrm{Vol}$.

The map $D^n \to \phi(D^n)$ is $n!$ to $1$ and, since $\phi$ is complex analytic, $\phi$ is orientation preserving. So
$$\int_{\phi(D^n)} \mathrm{Vol} = \frac{1}{n!} \int_{D^n} \phi^{\ast}(\mathrm{Vol}) = \frac{1}{n!} \int_{D^n} \det J^{\RR}_{\phi}$$
where $J^{\RR}_{\phi}$ is the Jacobian of $\phi$ considered as a smooth map $\RR^{2n} \to \RR^{2n}$. I will write $J^{\CC}_{\phi}$ when I instead want the $n \times n$ matrix of complex numbers coming from thinking of $\phi$ as a complex analytic map $\CC^n \to \CC^n$.

The relation between these two notions is this: $\det J^{\RR}_{\phi} = |\det J^{\CC}_{\phi}|^2$. (This is just linear algebra -- if $L/K$ is a degree $d$ field extension, $f: L^n \to L^n$ is a linear map and $g: K^{dn} \to K^{dn}$ is the linear map gotten by identifying $L$ with $K^d$, then $\det g = N_{L/K}(\det f)$.) We have the well known identity
$$\det J^{\CC}_{\phi}(z_1, \ldots, z_n) = \prod_{i<j} (z_i - z_j) = \sum_{w \in S_n} (-1)^w z_1^{w(1)-1} z_2^{w(2)-1} \cdots z_n^{w(n)-1}.$$
Here is the first reference I found.

So we need to compute
$$\frac{1}{n!} \int_{D^n} \sum_{u \in S_n} (-1)^u z_1^{u(1)-1} z_2^{u(2)-1} \cdots z_n^{u(n)-1} \overline{\sum_{v \in S_n} (-1)^v z_1^{v(1)-1} z_2^{v(2)-1} \cdots z_n^{v(n)-1}}.$$

We can distribute the product to get a sum of $(n!)^2$ terms of the form $\int_{D^n} \prod z_j^{a_j} \overline{z_j}^{b_j}$. If $a_j \neq b_j$, the integral on $z_j$ is zero, so we reduce to
$$\frac{1}{n!} \sum_{u \in S_n} \int_{D^n} (z_1 \overline{z_1})^{u(1)-1} (z_2 \overline{z_2})^{u(2)-1} \cdots (z_n \overline{z_n})^{u(n)-1}.$$
All $n!$ terms have the same integral, so we are reduced to the one integral
$$\int_{D^n} \prod (z_j \overline{z_j})^{j-1} = \prod_j \int_D (z \overline{z})^{j-1}.$$
Switching to polar coordinates,
$$\int_D (z \overline{z})^{j-1} = 2 \pi \int_{r=0}^1 r^{2j-1} dr = \frac{\pi}{ j}.$$
So the final answer is $\prod_{j=1}^n \tfrac{\pi}{j} = \tfrac{\pi^n}{n!}$.