Variance of a filtered signal
Your question (as expanded in the comments) is asking if we start with
$$
x(t)  = x_{\rm true}(t) + n_1(t)
$$
and filtering it using a filter $H(\omega)$ to get
$$
x_{\rm hp}(t) = h(t) * x(t) = x_{\rm true}(t) + n_2(t)
$$
where the variance of $n_1$ is $v_1$, then what is $v_2$, the variance of $n_2$?

That seems ill-posed because the filter $H$ will change $x_{\rm true}$ as well as the noise.

This page looks at the problem when $x_{\rm true}(t) = 0,\ \forall t$.

Here it is assumed that $n_1$ is white noise.

Where they get it is that
$$
r_{x_{\rm hp}}(l) = v_1 \cdot (h * h)(l)
$$
and $v_2$ is just this evaluated at $l=0$ i.e.
$$
v_2 = r_{x_{\rm hp}}(0)
$$

So you need to find the impulse response of your filter, convolve it with itself and find the zero-lag value of this.

As I said above, this doesn't quite answer your question but it might be a start.