Poles of an analog Gaussian filter
First of all, computing the poles of an (ideal) Gaussian filter is an impossible task, because its transfer function is not a rational function, and there are simply no poles. This is in contrast to the classic filter approximations, like Butterworth, Chebyshev, and Cauer filters, all of which have rational transfer functions, and, consequently, all of them have poles.

The frequency response of an ideal Gaussian filter is

$$H(\omega)=e^{-a\omega^2}\tag{1}$$

The corresponding transfer function is a function of $s=j\omega$:

$$G(s)=e^{-a(s/j)^2}=e^{as^2}\tag{2}$$

which is of course not rational, and which doesn't have any finite poles.

What you did is compute a rational approximation by developing $1/G(s)$ into a power series, and using only a finite number of terms. However, since you used $\omega$ as a complex variable instead of $s$, the poles are turned by $90$ degrees. This was already mentioned in the comments. The correct way of doing this is as follows:

$$1/G(s)=e^{-as^2}=\sum_{k=0}^{\infty}\frac{(-a)^k}{k!}s^{2k}\tag{3}$$

Using only a finite number of terms gives the following rational approximation of $G(s)$:

$$G(s)\approx\frac{1}{\sum_{k=0}^{n}\frac{(-a)^k}{k!}s^{2k}}\tag{2}$$

the poles of which are of course the roots of the denominator polynomial.