If we take the multiplicative group $G= \mathbb Q_p^\times$, then $\hat G \simeq \widehat{\mathbb Z} \times \mathbb Z_p^\times$ where the first factor is the (additive group of the) profinite completion of $\mathbb Z$ (by local CFT, this group is actually isomorphic to the Galois group of the maximal abelian extension $\mathbb Q_p^{ab} \mid \mathbb Q_p$ and hence of central interest). That first factor is well known to be isomorphic to the direct product of all additive groups of the $\ell$-adic integers for all (!) primes $\ell$, i.e. we get
$$\hat G \simeq \mathbb Z_p^\times \times\prod_{\ell \text{ prime}} \mathbb Z_\ell $$
Now I cannot shake a rigorous proof out of my sleeve right now, but I would be very surprised if this thing (well, the part $\prod_{\ell \text{ prime} \neq p} \mathbb Z_\ell$) is a $p$-adic Lie group.
On the other hand, I have a strong feeling that for compact $G$, we might be more lucky via the sources given in the comments.
Here is a completely (ha-ha) different way of describing the $p$-adic integers. Start with the ring of formal power series $\mathbf Z[[x]]$. Inside of here is the ideal $(x-p)$ of multiples of $x-p$. Take the quotient ring $R = \mathbf Z[[x]]/(x-p)$, a purely algebraic construction. This is isomorphic to the ring $\mathbf Z_p$!
The intuition here is that every $p$-adic integer is a power series in $p$ with coefficients in $\{0,1,\ldots,p-1\}$, so we could think about it as being a formal power series with restricted coefficients being evaluated at $p$. In $\mathbf Z_p$ an infinite series $\sum_{n \geq 0} a_np^n$ with $a_n \in \mathbf Z$ and no further restrictions at all will converge since the $n$th term tends to $0$ in $\mathbf Z_p$. Therefore numbers in $\mathbf Z_p$ might occur as a general integer power series evaluated at $p$ too.
Okay, define $\varphi \colon \mathbf Z[[x]] \to \mathbf Z_p$ by evaluation at $p$: $\varphi(f(x)) = f(p)$. This sends a formal power series $\sum_{n \geq 0} a_nx^n$ to the $p$-adically convergent series $\sum_{n \geq 0} a_np^n$, where the numbers $a_n$ are arbitrary integers. Properties of adding and multiplying convergent $p$-adic infinite series show $\varphi$ is a ring homomorphism. And $\varphi$ is surjective because every $p$-adic integer has a standard $p$-adic expansion with digits in $\{0,\ldots,p-1\}$.
It remains to show the kernel of $\varphi$ is the multiples of $x-p$.
The containment $(x-p)\subset \ker \varphi$ is immediate from $x-p \in \ker \varphi$. To prove $\ker \varphi \subset (x-p)$, suppose $f(x) = \sum_{n \geq 0} a_nx^n$ is in $\ker \varphi$, so $f(p) = 0$ in $\mathbf Z_p$. Then $a_0 \equiv 0 \bmod p\mathbf Z_p$, so the integer $a_0$ has to be an ordinary multiple of $p$: $a_0 = pb_0$ for some $b_0 \in \mathbf Z$. Then
$$
0 = f(p) = a_0 + a_1p + a_2p^2 + \cdots \equiv a_0 + a_1p \equiv pb_0 + a_1p \bmod p^2\mathbf Z_p,
$$
so $b_0 + a_1$ has to be an ordinary multiple of $p$: $b_0 + a_1 = pb_1$, so $a_1 = pb_1 - b_0$ for some integer $b_1$. Then show the condition $f(p) \equiv 0 \bmod p^3\mathbf Z_p$ implies $a_2 = pb_2 - p_1$ for some integer $b_2$. And so on: there is a sequence of integers $\{b_n\}$ such that
$$
a_0 = pb_0, \ \ a_n = pb_n - b_{n-1}
$$
for $n \geq 1$. That makes
$$
f(x) = \sum_{n \geq 0} a_nx^n = pb_0 + \sum_{n \geq 1} (pb_n - b_{n-1})x^n = (p-x)\sum_{n \geq 0} b_nx^n,
$$
so $f(x) \in (x-p)$ in $\mathbf Z[[x]]$. Thus $\mathbf Z[[x]]/(x-p) \cong \mathbf Z_p$ as rings for each prime $p$.
Remark. For different primes $p$, the rings $\mathbf Z_p$ are not isomorphic, so the rings $\mathbf Z[[x]]/(x-p)$ are nonisomorphic as $p$ varies. Note the contrast with $\mathbf Z[x]$: for each prime $p$, $\mathbf Z[x]/(x-p) \cong \mathbf Z$ as rings.
I did not use anything special above about $p$ being a prime number. If you've heard of $10$-adic integers, then the same argument shows $\mathbf Z_{10} \cong \mathbf Z[[x]]/(x-10)$.
Using the $x$-adic topology on $\mathbf Z[[x]]$, which makes $\mathbf Z[[x]]$ complete, the ring homomorphism $\varphi \colon \mathbf Z[[x]] \to \mathbf Z_p$ is continuous because the values at $p$ of all series in $x^k\mathbf Z[[x]]$ are very small for large enough $k$ (a group homomorphism of topological groups is continuous if and only if it is continuous at the identity), and continuity of $\varphi$ implies continuity of the induced mapping $\overline{\varphi} \colon \mathbf Z[[x]]/(x-p) \to \mathbf Z_p$ when using the quotient topology. This $\overline{\varphi}$ is also an open mapping (check first that $\varphi$ is an open mapping), so the bijection $\overline{\varphi}$ is a homeomorphism. Thus $\overline{\varphi}$ identifies $\mathbf Z[[x]]/(x-p)$ with $\mathbf Z_p$ both algebraically and topologically.
Best Answer
$\newcommand\Z{\mathbb Z}$
The argument you gave is correct, and if fact shows something a little more than you claim. You chose $x\in N$ of minimal valuation $n:=\nu_p(x)$. This means that $x=up^n$ where $u\in\Z_p^\times$ is a unit, so not only is $N\subset p^n\Z_p$, but in fact, $N=p^n\Z_p$. That is, these don't just form a cofinal system of finite quotients; there are exactly all finite quotients.For you bonus question, you certainly want to exclude $G$ itself since otherwise your completion will just give back the original group, i.e. $G\cong\varprojlim\limits_{\text{all }N\triangleleft G} G/N$ for completely formal reasons. Considering finite quotients has the benefit that the group you get in the end will be compact, since it will be a closed subset of the compact space $$\prod_{\substack{N\triangleleft G\\ [G:N]<\infty}}G/N,$$ where each $G/N$ is a finite space with discrete topology. Compact groups are especially nice, vaguely because compact spaces are nice and more concretely e.g. because you can do Fourier analysis on them. You can also make sense of the "order" of a profinite group (given as a supernatural number, e.g. the order of $\Z_p$ is $p^\infty$) and talk about Lagrange's theorem and Sylow $p$-subgroups and the like from finite group theory. Hopefully this helps to motivate profinite completions.
As a side note, it is sometimes useful to form the inverse limit of only some finite quotients; one could take the limit of only the cyclic quotients (and talk of procyclic groups), of only the quotients with $p$-power order (and talk of pro-$p$ groups), etc.
Finally, I don't think computing the profinite completion of $\Z$ is as trivial as you suggest. Let $\widehat\Z$ denote this completion. Can you show that $\widehat\Z\cong\prod_p\Z_p$?
Edit: I made blunder in the first paragraph. It is not the case that all finite index subgroups are of the form $p^n\mathbb Z_p$ (when I wrote "in fact, $N=p^n\Z_p$" I somehow had switched to thinking about ideals in my mind). So where does this leave the claim about cofinality?
We would like to say that the quotients $\Z_p/p^n\Z_p$ form a cofinal sequence of finite quotients of $\Z_p$, i.e. if $\Z_p/N$ is a finite quotient, then our projective system includes a morphism $\Z_p/p^n\Z_p\twoheadrightarrow\Z_p/N$, i.e. $p^n\Z_p\subset N$ for some $n$. This is the reverse inclusion of what you showed. Unfortunately, I have not been able to convince myself one way or the other about this system actually being cofinal, but I figured I should include this edit anyways just to make it clear that what I initially said was incorrect (See edit 2).
As a general note, one thing one might do to avoid these issues is to define the profinite completion of a topological group $G$ to be the inverse limit of its finite-index, open subgroups. Then is the same definition when $G$ is finite, and has the added benefit that it is obvious that the profinite completion of a profinite group is itself under this definition.
Edit 2: Thanks to @TorstenSchoeneberg in the comments for pointing out how to actually answer your question. If $N$ is a subgroup of finite index $n$, then $nx\in N$ for any $x\in\Z_p$, so $N\supset n\Z_p=p^{v_p(n)}\Z_p$ and we have cofinality.