Definition of the bias of an estimator I'm quite confused about the definition of the bias of an estimator.

Suppose we have unknown distribution $P(x, \theta)$, and construct the
estimator $\hat{\theta}$ that maps the observed data sample to values
that we hope are close to $\theta$. Then the bias of an estimator is
defined by:
$$Bias(\hat{\theta}) = \mathbb{E}(\hat{\theta}) - \theta ,$$

How can we calculate the $Bias(\hat{\theta})$ if we know neither the parent distribution $P(x, \theta)$ nor the true value of parameter  $\theta$?
Is there a need for any assumptions?

upd: suppose I've proved, that estimator of the mean ($\hat{\mu}=\frac{1}{m}\sum_{i=1}^mx_i$) is unbiased for the Gaussian distribution, but is there any distribution for which the same estimator is biased? Or is it possible to prove such an estimator is unbiased for any distribution?
I guess such a question is caused by the lack of intuition about estimator and its bias. May be there is some convenient way to think about it?

When we ask if an estimator is unbiased, it is important to add that it is unbiased for a given quantity $\theta$, so I would add that to the definition. On the calculation, you do the computations under the assumptions for the distribution you are working. It may have a parametric form or not.
To exemplify, let $F$ be a c.d.f. (playing the role of your $P(x, \theta)$), and assume we have an i.i.d. sample $(X_i)_{i=1}^n$ such that $X_i \sim F$. The only hypothesis we will assume is that $\theta = E[X_1] = \int xdF$ exists. Notice something very important: the distribution $F$  is not parametrized by $\theta$. Indeed, our estimators are not necessarily for "parameters", but rather for functions of your distribution (usually functionals). For example, there are problems in which you are interested in estimating the density of $F$ (the p.d.f.), and you do not think of it as a "parameter" of $F$.
Now, lets show that $\hat{\theta} = \frac{1}{n} \sum_{i=1}^n X_i$  is unbiased for $\theta$.
$$ E\left[\widehat{\theta}\right] = \frac{1}{n} E\left[\sum_{i=1}^n X_i\right] = \frac{1}{n} \sum_{i=1}^n E\left[X_i\right] = \frac{1}{n} \sum_{i=1}^n\int xdF = \theta \quad.$$
We only used properties of the expectation, which hold for any $F$, even if we do not know its "exact form".
Have we proved then that the sample mean is unbiased for any statistical model? Not at all. For instance, $\theta$ might not be well defined! If $F$ ranges over a Cauchy family, then it does not make sense to estimate $\theta$.
A similar problem might arise if the sample is not i.i.d.: $\theta$ may not be defining exactly what you want, and even if it is, the estimator might not be unbiased. For example, suppose that $(Y_i)_{i=1}^n$ are i.i.d. with $Y_i \sim \mathcal{N}(\theta, 1)$, but your sample is $\mathbf{X} = (X_i)_{i=1}^n$ with $X_i = (-1)^{n+1} Y_i$. If you want to estimate $\theta$, you actually have $\theta = E[X_1]$, so it is defined. However, the sample mean is not unbiased, as you can easily verify.
