$\newcommand{\v}{\operatorname{vol}}$The isoperimetric dimension is the maximum $d$ s.t.

$$\v(D)\leq C\cdot \v(\partial D)^{d/d-1}$$

for all open with smooth boundary $D\subset M$, differentiable manifold $M$, and universal constant $C$ for $M$. But it has been defined for graphs as well.

Q1: Let's start with metric measure space $(X,\Sigma, \mu, d)$, what are the minimal conditions and objects we need to add to it, to have the notion of isoperimetric dimension?

I think having just a measure space $(X,\Sigma, \mu)$ is not enough because we need a metric to define a perimeter length eg. via Minkoswki $\displaystyle \lim_{\varepsilon\to 0} \frac{\mu(A_{\varepsilon})-\mu(A)}{\varepsilon}$.

Q2: In other words, what is an example of a measure space $(X,\Sigma, \mu)$, and different metrics $d_1,d_2$ , each $(X,\Sigma, \mu, d_i)$ having different isoperimetric dimensions $\dim_1,\dim_2$

Finally,

Q3: Is there any counterexample for metric measure spaces, where we can't define isoperimetric dimension?

Probably some counterexample where there is no universal constant $C$ for any $d\in [0,\infty]$.

And preferably a continuum example. Because for example in the metric measure space $(\mathbb{Z}^{2},d_{discrete}, \mu_{counting })$ , sets are boundariless, but I think it is natural that the perimeter of say a 2D discrete square $Q$ is the number of edges between $Q$ and $Q^{c}$.

Whereas in the continuum I am more comfortable with perimeter of a set being

$$\displaystyle \lim_{\varepsilon\to 0} \frac{\mu(A_{\varepsilon})-\mu(A)}{\varepsilon}.$$