Consider the product of complex linear monic polynomials times polynomials of degree less than $n$, that is $\big( (z-\lambda), p(z)\big)\mapsto (z-\lambda)p(z)$. If we represent a polynomial by its coefficients, this produces a map $$h:\mathbb{C}\times\mathbb{C}^n\to \mathbb{C}^{n+1},$$ \begin{align} & \begin{bmatrix} \lambda \\ p_{0} \\ p_{1} \\ \vdots \\ p_{n-2} \\ p_{n-1} \end{bmatrix} \mapsto \begin{bmatrix} \phantom{p_0}-\lambda p_0, \\ p_0-\lambda p_1 \\ p_1-\lambda p_2\\ \vdots \\ p_{n-2}-\lambda p_{n-1}\\ p_{n-1} \phantom{-\lambda p_{n-1}} \end{bmatrix} \end{align} which is, by the Fundamental Theorem of Algebra, surjective onto the space of polynomials of positive degree at most $n$; in fact, a.e. with fiber of cardinality $n$. So the Gaussian measure of the image of $h$ with multiplicity (i.e., the integral of the multiplicity function $\# h^{-1}$) w.r.to Gaussian probability measure) is exactly $$n=\frac{1}{\pi^{n+1}} \int_{\mathbb{C}^{n+1}} \# h^{-1}(q)e^{-\|q\|^2} dq.$$ On the other hand, if we apply the Change of Variable formula, we find the following multiple integral with respect to the $2(n+1)$-dimensional Lebesgue measure on $\mathbb{C}\times\mathbb{C}^n\sim \mathbb{R}^{2(n+1)}$: $$\frac{1}{\pi^{n+1}}\int_{\mathbb{C}\times\mathbb{C}^n} \Big|\sum_{k=0}^{n-1}c_k\lambda^k\Big|^2e^{{-\sum_{k=0}^{n}| c_{k-1}-\lambda c_k|^2}} \mathrm d(\lambda,c_0,\dots,c_{n-1}),$$ where for the sake of notation we put $c_{-1}=c_n=0$ in the sum. If we compute directly this integral and find the value $n$, and use the obvious bound $\# h^{-1}\le n$ (meaning that a polynomial can't have more than $n$ roots), we have that the image of $h$ has full measure, so it is dense. Due to the elementary a priori bounds on the roots of a polynomial, we conclude that $h$ is indeed surjective onto the polynomials of positive degree, thus proving the Fundamental theorem of Algebra by means of multiple integrals.
The computation of the above integral turns out to be an elementary, yet not completely trivial exercise of iterated integrals and linear algebra (I will link the complete computation later). I wonder if there is a more clever way to perform the computation (of course, without using the FTA, nor any proof of it). Also, I am pretty sure -quia nihil sub sole novi, that somebody should already have done this computation, and I'd be glad to learn who.
edit Here is the elementary computation, for this question, and to whom is interested.