~~The simplest~~ `Another`

group answering the question is the Heisenberg group (over $\mathbb{Z}$):
$$
H_3(\mathbb{Z}) = \left\lbrace \left(\begin{smallmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{smallmatrix}\right) \bigg| x, y, z \in \mathbb{Z} \right\rbrace
$$
This group is generated by two elements (the matrices $\left(\begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{smallmatrix}\right)$ and $\left(\begin{smallmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right)$ (since the matrix $\left(\begin{smallmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix}\right)$ is the commutator of the two other.

On the other hand it has volume growth of the type $V(n) \simeq n^4$ and satisfies a 4-dimensional isoperimetric inequality. So this answers the first question.

As for the second question, note first that your inequality implies a lower bound on the volume growth of the type $n^d$. Indeed, by looking at $B_n$ the ball of radius $n$, one gets that $|\partial B_n| \geq C|B_n|^{1-1/d}$. Since the vertices are of bounded degree, $|B_n| -|B_{n-1}| \geq C'|B_n|^{1-1/d}$. The easiest way to conlcude is to introduce a piecewise affine extension $b(x)$ of the function $n \mapsto |B_n|$. Then your inequality reads $b'(x) \geq C'b(x)^{1-1/d}$ which integrates to a bound of $b(x)^{1/d} \geq C'' x$

On the other hand there is a reverse inequality (see Coulhon, Thierry, and Saloff-Coste, Laurent. "Isopérimétrie pour les groupes et les variétés.." Revista Matemática Iberoamericana 9.2 (1993): 293-314. There is a survey written in English and availiable online by Pittet and Saloff-Coste, see here. Look at section 1 (the function you are interested in is $\tfrac{1}{J}$, but not $J$ or $I$) and section 7 (more precisely Theorem 7.0.10): one has that $V(n) \succeq n^d$ implies $J(t) \preceq t^{1/d}$ (which implies $\tfrac{1}{J(t)} \succeq t^{-1/d}$, i.e. a $d$-dimesional inequality).

So this means the answer to your second question is: the isoperimetric dimension of a group is an integer or $+\infty$ (~~see Yves' answer for an example of a group with infinite dimension which is amenable [in fact solvable]~~ `a group which is amenable [in fact solvable] and has infinite isoperimetric dimension is`

$\mathbb{Z}^\infty \rtimes \mathbb{Z}$`, basically the limit as`

$k \to \infty$`of the groups in Yves' answer`

).

Indeed, if a group had an isoperimetry of $k +\epsilon$ for some $k \in \mathbb{N}$ and $\epsilon \in ]0,1[$ then, its growth would be at least $k+\epsilon$. Since the growth exponent is an integer, then its growth is actually at least $k+1$. In turns this implies the group has a $(k+1)$-dimensional isoperimetry.