You can see $\zeta^*$ as generated by entourages of the form $ U^f_\varepsilon = \{ (n, m) | \| f(n) - f(m) \|_\infty < \varepsilon \} $
for all $\varepsilon > 0$, $n \in \mathbb{N}$, and $f : \mathbb{N} \rightarrow [0, 1]^n$.
Using the same argument, you should be able to prove that given $\varepsilon$ and $f : \mathbb{N} \rightarrow [0, 1]^n$, there exists a finite $U^f_\varepsilon$-net in $\mathbb{N}$ (you only need to use the fact that $[0, 1]^n$ itself is totally bounded).
Since now the $U^f_\varepsilon$ constitute a fundamental system of entourages for $\zeta^*$, it follows that $\zeta^*$ is totally bounded.
For your second question, I assume you mean "all continuous functions". Since $\mathbb{Q}$ is a metric space, you get the uniformity induced by the metric, which is of course not totally bounded.
EDIT: Oops, this is not true.
To see that, take a ball $B$ of radius $\varepsilon$ centered in $x_0$, then (using the above notation), $U^f_{1/2}(x_0) \subseteq B$, where $f$ is a function such that $f(x_0) = 1$ and $f \equiv 0$ on the complement of $B$.
EDIT: This proves that the initial uniformity induces the Euclidean topology. The initial uniformity is still totally bounded by the same argument as for $\mathbb{N}$.
Now, coming to your real question, using all uniformly continuous functions to generate a uniformity on $\mathbb{Q}$ still yields a totally bounded uniformity (since it is coarser than the initial uniformity). The point is that this uniformity $\mathcal{U}$ does not coincide with the one induced by the metric (let's call it $\mathcal{E}$), because you are limiting yourself to bounded functions. So all three uniformities are distinct, and have the following inclusion relationships:
$ \mathcal{U} \subset \zeta^*,\quad \mathcal{U} \subset \mathcal{E} $
even though they all induce the same topology.