I don't know if you can construct a sequence as you've proposed above, but I do know that the inequality you proposed is sharp. Proof of this is neatly presented in Becker's "Inequalites in Fourier Analysis'' (1975):
http://www.jstor.org/stable/1970980
The sharp Young's inequality reads:
$$\|f \ast g \|_r \leq \left( A_p A_q A_{r'} \right)^n \|f\|_p\|g\|_q$$
where $A_s = \left( \frac{s^{1/s}}{s'^{1/s'}} \right)$, $\frac{1}{s} + \frac{1}{s'} = 1$, $f$ and $g$ are functions on $n$ dimensions, and $1 \leq p,\,q,\,r\leq \infty$, as usual. In order to show that your inequality is sharp, it's only necessary to note that $A_s A_{s'}=A_1=1$.
This inequality is saturated whenever $f$ and $g$ are both Gaussians, according to the paper.
Q2: I'm not an expert on this, so I would defer to someone else. Regarding the definition, "group" here is referring to a "topological group"; see Locally compact group for details and examples. I think Ball is implicitly referring to convolutions with respect to Haar measures on such spaces, but I'm not sure.
Q3, Q8: If you glance at the Brascamp and Lieb paper, you can infer that the constant Ball is referring to is basically the $c$ in $\|f * g\|_s \le c \|f\|_p \|g\|_q$. Presumably one is able to prove the inequality holds in general when $c=1$, but one might ask if you can prove the inequality for a smaller value of $c$. Ball notes that spaces where constant functions are integrable, equality is attained ($\|f*g\| = \|f\|_p \|g\|_q$), so you can't make $c$ smaller than $1$ in general. But in the specific case of $\mathbb{R}$, apparently the inequality holds for a smaller $c$.
Q4: For a function $u \in L^s$ and $h \in L^r$ with $1/s + 1/r=1$, Hölder's inequality $\int u(x)h(x) \, dx \le \|u\|_s \|h\|_r$ attains equality when $|u|^s$ and $|h|^r$ are equal almost surely up to a multiplicative constant. In particular, if $h$ is of the form $h = c u^{s/r}$ where $c = \|u^{s/r}\|_r^{-1}$ so that $\|h\|_r = 1$, then we have $\int u(x) h(x) \, dx = \|u\|_s$. Ball is applying this with $u=f*g$.
Response to comment: $\|h\|_r=1$ is only necessary for Ball's claim that "$\|f*g\|_s$ can be realized as $\int (f*g)(x) h(x) \, dx$ with $\|h\|_r=1$." It is not necessary for the subsequent inequalities that keep $\|h\|_r$ on the right-hand side. The purpose is to show the equivalence between the original convolution inequality and the modified form (you can see that the Brascamp and Lieb paper considers the second form).
Q5, Q6: The convolution integral is complicated because of the $g(x-y)$ term which involves both dummy variables $x$ and $y$ being integrated. I guess it is easier to to consider an integrand $f(u) g(v) h(w)$ that decomposes with respect to the dummy variables (each factor only involves one variable). Note that this isn't a "free lunch" scenario where we have magically untangled the dummy variables; the dummy variables are still intertwined by the subspace condition $u+v+w=0$. But this conversion is helpful for the purpose of getting intuition on why convolutions are relevant in estimating volumes (see the sentence describing the "area of a slice of a cube"), which is really the purpose of these few pages.
You are correct that the Jacobian factor would matter for determining the constant mentioned above, but I think Ball is just trying to give intuition here.
Q7: I think any measure that assigns measure to a set proportional to its surface area works. Exactly which measure this is (i.e., what is the multiplicative constant that converts from area to measure) would depend on an explicit parameterization of the plane, and I think the constant would just be the Jacobian constant (or could get absorbed into it). Since these measures are all proportional up to a constant, I don't think it matters for the sake of the discussion which only concerns intuition.
Response to comment: I did not really imply what you wrote. In the case of the plane $H$ defined in the text, I was referring to measures of the form $\mu(A) \propto \text{area}(A)$ for $A \subseteq H$. (No $\partial A$, and not $\mathbb{R}^n$.) In the context of the convolution inequality, this just comes out of the transformations. In terms of the broader context, I didn't read the rest of the text, but I guess there is some geometric context where you are interested in areas/volumes, so such measures (that are proportional to areas/volums) would be relevant.
Best Answer
Yes, Young's inequality can be shown to hold for arbitrary locally compact groups — under suitable integrability assumptions on $f$ and $g$, see Hewitt–Ross, Abstract Harmonic Analysis, I, Theorem (20.18) on page 296 for the precise statement.
If $G$ happens to be abelian, compact, discrete (or, more generally, unimodular) then these assumptions translate to: If $f \in L^{p}$, $g \in L^q$ and $\frac{1}{p} + \frac{1}{q} = 1 + \frac{1}{r}$ for $1 \leq p, q, r \leq \infty$ then $f \ast g \in L^r$, and
$$\|f \ast g\|_r \leq \|f\|_p\,\|g\|_q.$$
Replacing integrals by sums robjohn's argument here carries over painlessly to $\mathbb{Z}$ or $\mathbb{Z}^d$.