In general terms my estimator is $\sum_{i=1}^n {\omega_i * x_i \over \sum _{i=1}^n {\omega_i}}$ where $x_i$ are realizations of the instrumental r.v., $w_i$ its corresponding importance weighs and $n$ the sample size.
What is the variance of this estimator?
This is some work showing the Delta Method for approximating the variance of a ratio.
Let $X_1, \ldots, X_n \overset{iid}{\sim} q()$ be samples from your normalized instrumental density $q(\cdot)$. Let $p(\cdot) = C^{-1}p_u(\cdot)$ be your target density. Assume you can only evaluate $p_u$. Call $w_i = w_i(x_i) = p_u(x_i)/q(x_i)$.
The Delta Method is justified with Taylor approximations. Call $A = \frac{1}{n}\sum_i w_i x_i$, $B=\frac{1}{n}\sum_j w_j$, the numerator and denominator of your expression. Also, call $\mu_A$ and $\mu_B$ their expected values. That's
$$ \sum_{i=1}^n {w_i * x_i \over \sum _{i=1}^n {w_i}} = \frac{A}{B}. $$
Delta Method takes the Taylor approximation,
$$ f(A,B) \approx f(\mu_A,\mu_B) + f_{A}(\mu_A,\mu_B)(A-\mu_A) + f_B(\mu_A,\mu_B)(B-\mu_B) $$
and takes the variance on both sides:
$$ \text{Var}\left[\frac{A}{B}\right] \approx [f_{A}(\mu_A,\mu_B)]^2\text{Var}[A] + [f_B(\mu_A,\mu_B)]^2\text{Var}[B] + 2f_{A}(\mu_A,\mu_B)f_B(\mu_A,\mu_B)\text{Cov}(A,B). $$ Or in your case:
\begin{align*} &\frac{1}{\mu_B^2}\frac{1}{n}E[(WX - \mu_A)^2] + \frac{\mu_A^2}{\mu_B^4}\frac{1}{n}E[(W - \mu_B)^2] - 2\frac{1}{\mu_B}\frac{\mu_A}{\mu_B^2}E[W^2X] + 2\frac{1}{\mu_B}\frac{\mu_A}{\mu_B^2}E[WX]E[W] \\ &= \frac{1}{\mu_B^2}\frac{1}{n}\left\{E[W^2X^2] + \frac{\mu_A^2}{\mu_B^2}E[W^2] - 2 \frac{\mu_A}{\mu_B}E[W^2X] \right\} \\ &= \frac{1}{\mu_B^2}\frac{1}{n}\left\{ E[(XW - \frac{\mu_A}{\mu_B}W)^2]\right\}\\ &= \frac{1}{\mu_B^2}\frac{1}{n}\left\{ E[W^2(X - \frac{\mu_A}{\mu_B})^2]\right\}, \end{align*} where we use an uppercase $W$ to denote any of the random unnormalized weights. If you plug in the sample estimates for all the above quantities you get
$$ \frac{1}{n}\frac{\frac{1}{n}\sum_i w_i^2(x_i - A/B)^2 }{B^2 } = \sum_{i=1}^n \left[\frac{w_i}{\sum_j w_j}\right]^2(x_i - A/B)^2. $$
I used this as a reference: http://statweb.stanford.edu/~owen/mc/Ch-var-is.pdf
A very crude approximation to the variance of the self-normalised importance sampling estimator $$ \hat{\mu}_n = \sum_{i=1}^n {\omega_i x_i \over \sum _{i=1}^n {\omega_i}} $$ is $$ \textrm{Var}_q(\hat{\mu}_n) \approx \textrm{Var}_p(\hat{\mu}_n) (1 + \textrm{Var}_q(W)). $$ where $p$ is the target distribution and $q$ the importance distribution.
To get a more accurate estimate than the Delta Method (shown in answer by @Taylor), use bootstrapping https://en.wikipedia.org/wiki/Bootstrapping_(statistics) and https://www.crcpress.com/An-Introduction-to-the-Bootstrap/Efron-Tibshirani/p/book/9780412042317. As a bonus, that will give you an estimate of the entire distribution.
However if you want to get an estimate of variance without (prior to) having any data, then use the Delta Method.
