At the start of this document http://www.math.lsa.umich.edu/~lagarias/575chomework/p-adic-chap5.pdf Lagarias draws the analogy between writing any integer in base $p$, and writing any polynomial in the base of some linear polynomial $X-\alpha$ where $\alpha$ is a complex number.
The prime integers are ordered, and when writing in base $p$ every digit $ a_i$ fulfils $0\leq a_i\leq p-1$. Is there any analogous ordering of the linear polynomials $(X-\alpha)$, and therefore any measure by which $0\leq a_i\leq (X-\alpha)$?
What you are looking for is the notion of a Euclidean domain. Roughly speaking, a Euclidean domain is a domain $R$ which has a function $d:R\setminus\{0\}\to\mathbb{N}$ (measuring the "size" of elements of $R$) such that whenever you divide by an element $b\in R$, you get a remainder that is "smaller" than $b$. More precisely, if $b\in R$ is nonzero and $a\in R$, then there exist $q,r\in R$ (the "quotient" and "remainder" of dividing $q$ by $r$) such that $a=bq+r$ and either $r=0$ or $d(r)<d(b)$.
For integers, you take $d(n)=|n|$, and this is just the statement that when you divide with remainder by a positive integer $b$, you get a remainder $r$ which satisfies $0\leq r<b$. This relates to what you said because the "base $p$ digits" are exactly the possible remainders you can get when dividing an integer by $p$.
For polynomials, you take $d(f)$ to be the degree of $f$. So $d(X-\alpha)=1$ for all $\alpha$, while $d(a_i)=0$ for any constant $a_i$. In this sense, $a_i$ is "smaller" than $X-\alpha$. If you like, you can make this into a total preorder of all the nonzero polynomials by saying $f\leq g$ iff $d(f)\leq d(g)$. This preorder cannot distinguish between polynomials of the same degree, however.
(If you want a genuine total order of the polynomials, there won't be any natural such ordering. You could of course simply pick any order you like by which all constants are less than all higher degree polynomials, but there isn't any canonical choice of such an order. In particular, there is no order compatible with the ring structure in the sense of an ordered ring, since that would restrict to the constants to make $\mathbb{C}$ an ordered field, which is impossible.)