Hypothesis testing. Why center the sampling distribution on H0? A p-value is the probability to obtain a statistic that is at least as extreme as the one observed in the sample data when assuming that the null-hypothesis ($H_0$) is true.
Graphically this corresponds to the area defined by the sample statistic under the sampling distribution which one would obtain when assuming $H_0$:

However, because the shape of this assumed distribution is actually based on the sample data, centering it on $\mu_0$ seems like an odd choice to me. 
If one would instead use the sampling distribution of the statistic, i.e. center the distribution on the sample statistic, then hypothesis testing would correspond to estimating the probability of $\mu_0$ given the samples.

In that case the p-value is the probability to obtain a statistic at least as extreme as $\mu_0$ given the data instead of the above definition.
Additionally, such an interpretation has the advantage of relating well to the concept of confidence intervals:
A hypothesis test with significance level $\alpha$ would be equivalent to checking whether $\mu_0$ falls within the $(1-\alpha)$ confidence interval of the sampling distribution.

I thus feel that centering the distribution on $\mu_0$ could be an unneccessary complication. 
Are there any important justifications for this step which I did not consider? 
 A: Suppose $\boldsymbol X = (X_1, X_2, \ldots, X_n)$ is a sample drawn from a normal distribution with unknown mean $\mu$ and known variance $\sigma^2$.  The sample mean $\bar X$ is therefore normal with mean $\mu$ and variance $\sigma^2/n$.  On this much, I think there can be no possibility of disagreement.
Now, you propose that our test statistic is $$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim \operatorname{Normal}(0,1).$$  Right?  BUT THIS IS NOT A STATISTIC.  Why?  Because $\mu$ is an unknown parameter.  A statistic is a function of the sample that does not depend on any unknown parameters.  Therefore, an assumption must be made about $\mu$ in order for $Z$ to be a statistic.  One such assumption is to write $$H_0 : \mu = \mu_0, \quad \text{vs.} \quad H_1 : \mu \ne \mu_0,$$ under which $$Z \mid H_0 = \frac{\bar X - \mu_0}{\sigma/\sqrt{n}} \sim \operatorname{Normal}(0,1),$$ which is a statistic.
By contrast, you propose to use $\mu = \bar X$ itself.  In that case, $Z = 0$ identically, and it is not even a random variable, let alone normally distributed.  There is nothing to test.
A: From what I gather, you are arguing that it makes more sense to 'flip' $H_0$ and $H_1$. 
I find it helpful to think of hypothesis testing as a proof by contradiction. We assume $H_0$ to be true, then show that evidence indicates such an assumption is flawed, thus justifying the rejection of $H_0$ in favor of $H_1$. 
This works because when we assume $H_0$ and center our distribution there, we can determine how likely/unlikely our observation is. For example, if $H_0: \mu = 0$ vs.  $H_1: \mu \neq 0$ and we determine from our testing that there is a less that 5% chance that the true mean $\mu$ actually equals 0, we can reject $H_0$ with 95% confidence. 
The reverse is not necessarily true. Say we do an experiment and determine that there is actually a 30% chance that the null hypothesis still holds. We cannot reject the null, but we also do not accept it. This situation does not show that $H_0$ (the null) is true, but that we do not have the evidence to show that it is false.
Now imagine if we flipped this situation. Say we assume $H_1$ and find that given our results, the likelihood of $H_0$ is 5% or less, what does that mean? Sure we can reject the null, be can we necessarily accept $H_1$? It is hard to justify accepting the thing we assumed to be true in the beginning. 
Showing that $H_0$ is false is not the result we are after; we want to argue in favor of $H_1$. By doing the test in the way you describe, we are showing that we do not have evidence to say that $H_1$ is false, which is subtly different than arguing $H_1$ is true. 
A: 
However, because the shape of this assumed distribution is actually based on the sample data, centering it on H0 seems like an odd choice to me. 

This is actually not true. The shape of this assumed distribution comes from accepting $H_0$ as true. Sample is not directly involved in that, other than by some assumptions. Using the sample directly, is not enough. You need also the null hypothesis to hold.

If one would instead use the sampling distribution of the statistic, i.e. center the distribution on the sample statistic, then hypothesis testing would correspond to estimating the probability of H0 given the samples.

The question is: how do you estimate a probability of something which you assume is true. In our case if you assume $H_0$ as true, is futile to try to estimate the probability that $H_0$ is true.

I thus feel that centering the distribution on H0 is an unneccessary complication. 

You don't have two distributions there, there is only one, the one assumed to be your ground truth, aka the one which comes with $H_0$. There is however a sampling distribution derived from sample, but this is not involved in the hypotheses you use.
I good exercise would be to try to replicate the same logic with an asymmetric distribution. Take chi-square distribution like in chi square independence test. Are you able to reproduce it? I think the answer is no.
