In particular, then, you want a total ordering on $\mathbb{Q}[\omega]$ for some $p$-adic $\omega$, with $\omega$ greater than the elements of $\mathbb Q$ in the order.
If $\omega$ is transcendental over $\mathbb Q$ in $\mathbb Q_p$, then $\mathbb Q[\omega] \cong \mathbb Q[x]$, where $x$ is an indeterminate, so you'd want a total order on the ring of polynomials over $\mathbb{Q}$ in which $x>r$ for all rational $r$
But if the total order is compatible with multiplication and addition, we have the following:
$$\forall n\geq 0, r\in \mathbb Q: x^{n+1}>rx^n$$
From this, we can deduce that a polynomial $p(x)$ is greater than zero if it's first coefficient is greater than zero.
That entirely defines the total order on $\mathbb Q[x]$. $p(x)>q(x)$ if the leading coefficient of $p(x)-q(x)$ is positive.
If $\omega$ is not transcendental, then you can show there can be no such ordering, basically by the same argument that $\omega^n$ must be strictly greater than any linear combination of smaller powers of $\omega.$
So what this shows is that the only numbers that you can consider "infinites" are transcendentals, but that any attempt to understand infinite values this way is more a study of $\mathbb{Q}[x]$ rather than the $p$-adics.
$\mathbb{Z}$ itself is Archimedean, but it is not divisible: for example, there is no $b \in \mathbb{Z}$ such that $2b = 1$.
The ring $Q[x]$, with the total ordering in which a polynomial in $x$ with rational coefficients is positive if and only if its trailing coefficient is positive, is divisible, but it is not Archimedean: for example, $0 < nx < 1$ for every positive integer $n$, because $1 - nx > 0$.
Further to my comment above, in response to the revised question,
these extracts are from F. Loonstra, "Ordered groups",
Proc. Nederl. Akad. Wetensch. 49 (1946), pp. 41-46 (PDF
here):
If an ordered set $O$ is dense in itself, then generally $O$ need
not be dense-ordered. (A set $O$ is called dense-ordered if for
each pair of elements $a$ and $b$ of $O$ there is at least one
element $c$ "between" $a$ and $b$.) For ordered groups however we
prove:
If $G$ is an ordered group dense in itself, then $G$ is densely
ordered.
If there was namely between two elements $a$ and $b$ no further
element, then every element had an immediate successor and
predecessor and thus $G$ was not dense in itself. Conclusion: An
ordered group $G$ is dense in itself if and only if $G$ is densely
ordered. If on the contrary one element is isolated, then every
element is necessarily isolated; consequently:
An ordered group $G$ is either a discretely ordered group, which
means that every element has an immediate successor and predecessor,
or a densely ordered group. The property of an ordered group as
being discrete resp. dense is as a topological property invariant
for a mapping preserving the order.
If we divide an ordered set $N$ into two subsets $N_1$ and $N_2$
with the conditions:
I. Every element of $N$ belongs to one, but only one of the sets
$N_1$ and $N_2$;
II. Neither $N_1$ nor $N_2$ is empty;
III. All elements of $N_1$ precede those of $N_2$.
then we call this division of $N$ a Dedekind-cut $N_1 | N_2$ in $N$.
There are only the following cases that exclude each other:
$1$. $N_1$ has a last, $N_2$ a first element; then we call the
cut $N_1 | N_2$ a sault in $N$.
$2^a$. $N_1$ has a last, $N_2$ no first element;
$2^b$. $N_1$ has no last, $N_2$ has a first element; we call
the cut in $2^a$ and $2^b$ a continuous cut.
$3$. $N_1$ has no last. $N_2$ has no first element; the cut is
called a lacuna in the set $N$.
The ordered set $N$ is called continuously ordered, if no cut in $N$
defines a sault or a lacuna.
A non-archimedean ordered group is not continuously ordered.
If the ordering was continuous then of every cut $A | B$ either $A$
had a last, $B$ no first element, or just the reverse. We shall try
to prove, however, that in every non-archimedean ordered group there
is to define a non-continuous cut. Suppose $e < a < b$ and $a$
non-archimedean with regard to $b$. Let the class $A$ contain all
elements $\leqslant a^n$ for a positive integer $n$, $B$ the
remaining elements; $B$ is not empty, as $b \in B$. $A$ has no last
element, nor has $B$ a first element, for $B$ also contains the
elements $ba^{-1} > ba^{-2} > \cdots$ elements that precede $b$.
Therefore a non-archimedean ordered group is never a continuous
ordered set or directly: continuously ordered groups are necessarily
archimedean ordered groups.
A densely ordered group $G$ has the property that to every element
$a > e$ there exists an element $b > e$, so that $b^2 < a$.
[A proof of that is provided, but ...]
Without a proof we pronounce the following theorem:
In a continuous ordered group there exists to every element $a > e$
one element $b > e$ with $b^n = a$.
[I was just typing this out, making edits and corrections, and joining
it all up, and didn't notice that he hadn't provided a proof of this part!
I think I have a proof in my own handwritten notes from years ago, but
it might not be easy to find, and may be wrong. Let's just hope the result
is obvious ... er ... an exercise for the reader?] :)
Noah Schweber's answer to the following question may help to fill in the, er, gap
I've just left:
Is $\mathbb{R}$ the only complete ordered Abelian group?.
Best Answer
EDIT: I thought it would be appropriate, given the perhaps unexpected descriptive set-theoretic nature of this answer, to give a foreword explaining the focus on the Baire property (BP for short). A subset of a topological space has the BP if it differs from an open set by a meager set (a set contained in the union of countably many nowhere dense sets). In the setting of a Polish space (such as $\mathbb{Z}_p$ or the real numbers), the BP sets contain the Borel sets (and continuous images of Borel sets) and satisfy some nice regularity properties. In this context BP sets should be considered analogous to measurable sets, with meager sets serving as analogs of the null sets. For example, the Polish space itself is not meager, and the Kuratowski-Ulam theorem asserts that a subset of the plane is meager if and only if only a meager set of vertical sections are nonmeager (this is basically a Fubini theorem).
One particularly noteworthy fact is that it's consistent with the axioms $\mathtt{ZF+DC}$ that every subset of a Polish space has the BP. Recall that $\mathtt{ZF}$ is the standard axiomatization of set theory without choice, and $\mathtt{DC}$ is the axiom of dependent choice. Intuitively, $\mathtt{DC}$ gives you the freedom to make a countable sequence of choices, where each choice is allowed to refer to properties of your previous choices (so they aren't "independent"). Dependent choice is enough to do almost all of common mathematics: you can perform most of analysis, carry out typical inductive constructions, Borel sets behave reasonably, the first uncountable cardinal is not a countable union of countable sets, etc. Some contexts in which $\mathtt{DC}$ doesn't bestow the full power of $\mathtt{AC}$ include performing wildly nonconstructive acts like building a Vitali set or choosing bases from huge vector spaces. So, I think that $\mathtt{ZF+DC}$ is a reasonable framework in which to carry out your request for an "explicit" linear order of $(\mathbb{Z}_p,+)$. Once we rule out the existence of such a linear order with the property of Baire, we're therefore forced to concede that no argument producing this order may be carried out in $\mathtt{ZF+DC}$, dashing our hopes of an explicit construction.
By the way, I focus on $(\mathbb{Z}_p, +)$ rather than $(\mathbb{Q}_p,+)$ simply for convenience. It should be clear that any order of the latter induces an order of the former, so if anything the problem is harder for $\mathbb{Z}_p$.
There is no linear order of the additive group $(\mathbb{Z}_2,+)$ of $2$-adic integers which has the property of Baire (with respect to the usual Polish topology). In particular, it is consistent with $\mathtt{ZF+DC}$ that no such order exists at all, so a large fragment of the axiom of choice is indeed required to build such an order. Analogous arguments will work for all $\mathbb{Z}_p$, with slightly more obnoxious notation.
From here on we will identify elements of $\mathbb{Z}_2$ with infinite binary strings, that is, elements of $2^\omega$ with the product topology. We define an equivalence relation $E_0$ on $2^\omega$ by setting two strings equivalent iff they differ in finitely many coordinates. This $E_0$ has a nice interpretation in $\mathbb{Z}_2$: $x$ and $y$ are $E_0$ related iff their difference is a (standard) integer. More precisely, $x E_0 y$ iff for some $n$, $x + 1 + 1 + \cdots + 1 (n \mbox{ times}) = y$ or vice-versa, where $1$ denotes the standard integer $1$ (i.e., the sequence $10000\ldots$).
(That last part isn't literally true, since the constant $1$ sequence plus $1$ equals the constant $0$ sequence. But it is true off of the eventually constant sequences, which is enough to make the below argument go through (since there are only countably many eventually constant sequences).)
We use without proof two standard facts about $E_0$:
Now, given a putative order $<$ with the BP, we can partition $2^\omega$ into three $E_0$-invariant BP pieces:
(here $0$ is the identity element of the group: the constant $0$ sequence). So $X_-$ is the union of the $E_0$-classes which are entirely negative, $X_+$ the union of those entirely positive, and $X_0$ the union of those which are sometimes positive and sometimes negative.
(Technically, these pieces might not have the BP, but by Kuratowski-Ulam there's some element in $2^\omega$ we can use in place of $0$ to make the pieces have the BP. For ease of notation, let's assume $0$ works.)
We first observe that $X_0$ is meager. Note that the set $\{x : 0 \leq x < 1\}$ (which is meager by Fact 1) hits each $X_0$ class in exactly one point, so $X_0$ is the union of countably many homeomorphic translations (namely, the standard integer shifts) of a meager set, thus is meager.
Now let $f: 2^\omega \to 2^\omega$ denote the bitflipping homeomorphism, so $f(01001110\ldots) = 10110001\ldots$. We note that $x \in A_-$ iff $f(x) \in A_+$, since $x + f(x) + 1 = 0$ for all $x$. This means that $A_-$ cannot be comeager, else $A_+ = f[A_-]$ would be a disjoint comeager set. But then by Fact 2, $A_-$ is meager, thus so is $A_+ = f[A_-]$, and consequently we've written $2^\omega$ as the union of three meager sets. So we've hit a contradiction.
(*) By request, here is a reference for Fact 2: Theorem 3.2 of G. Hjorth: Classification and Orbit Equivalence Relations, Mathematical Surveys and Monographs, 75, American Mathematical Society, Providence, RI, 2000. Although actually this theorem is overkill for this special case -- here's a sketch of a more elementary argument that works here.
Suppose that $B \subseteq 2^\omega$ is nonmeager; we want to show that $[B]_{E_0} = \{x: \exists y \in B\ (x E_0 y)\}$ is comeager. By localization, there's a basic open set $U$ such that $B \cap U$ is comeager in $U$. We can find a finite binary string $s$ let's say of length $n$ such that $U$ contains all elements of $2^\omega$ beginning with $s$. Now look at the $2^n$ homeomorphisms of $2^\omega$ which flip some subset of the first $n$ bits of a string and leave the rest unchanged. These maps send each $x$ to something $E_0$-related to $x$, so it follows that $[B]_{E_0}$ is comeager in the union of $U$'s images under these maps. But the union of these images is all of $2^\omega$!