Hint:
Note that the Euclidean norm is a particular case of a $p$-norm and for these norms the triangle inequality can be proved using the Minkowky inequality.
Anyway, the Euclidean norm is the only $p$-norm that satisfies the parallelogram identity ( see: Determining origin of norm), so it is coming from an inner product.
About the addendum.
In an $n$ dimensional real space we can prove the C-S inequality with simply algebraic methods (see here). So, yes, in this case we can proof the triangle inequality without explicitly using an inner product space.
You can define a euclidean norm for any basis: take a basis $b = \{b_1, b_2\}$ for $\mathbb{R}^2$ and define $\|\cdot\|_{2,b} : \mathbb{R^2} \to \mathbb{R}$ as:
$$\|x\|_{2,b} = \|\alpha b_1 + \beta b_2\|_{2,b} = \sqrt{\alpha^2 + \beta^2}$$
You can verify that $\|\cdot\|_{2,b}$ is a norm. Indeed, these norms are different for different bases (they are all eqiuvalent, though).
However, for any fixed basis $b$, its definition is basis independent since you can always write a formula for $\|\cdot\|_{2,b}$ using coordinates only:
Let $b = \{(x_1, y_1),(x_2, y_2)\}$ be a basis.
Then we have:
$$(x, y) = -\frac{x_2y+xy_2}{-x_2y_1+x_1y_2}\,(x_1,y_1)+\frac{x_1y-xy_1}{-x_2y_1+x_1y_2}\,(x_2,y_2)$$
So $$\|(x,y)\|_{2,b} = \sqrt{\left(\frac{x_2y+xy_2}{-x_2y_1+x_1y_2}\right)^2 + \left(\frac{x_1y-xy_1}{-x_2y_1+x_1y_2}\right)^2}$$
which is entirely basis independent.
Best Answer
There is a "standard" way to consider normed spaces over arbitrary fields but these are not well-behaved in the case of scalars in finite fields. If you want to work with norms on vector spaces over fields in general, then you have to use the concept of valuation.
Valued field: Let $K$ be a field with valuation $|\cdot|:K\to\mathbb{R}$. This is, for all $x,y\in K$, $|\cdot|$ satisfies:
The set $|K|:=\{|x|:x\in K-\{0\}\}$ is a multiplicative subgroup of $(0,+\infty)$ called the value group of $|\cdot|$. The valuation is called trivial, discrete or dense accordingly as its value group is $\{1\}$, a discrete subset of $(0,+\infty)$ or a dense subset of $(0,+\infty)$. For example, the usual valuations in $\mathbb{R}$ and $\mathbb{C}$ are dense valuations. The valuation is said to be non-Archimedean when it satisfies the strong triangle inequality $|x+y|\leq\max\{|x|,|y|\}$ for all $x,y\in K$. In this case, $(K,|\cdot|)$ is called a non-Archimedean valued field and $|n1_K|\leq1$ for all $n\in\mathbb{Z}$. Common examples of non-Archimedean valuations are the $p$-adic valuations in $\mathbb{Q}$ or the valuations of a field that is not isomorphic to a subfield of $\mathbb{C}$.
Norm: Let $(K,|\cdot|)$ be a valued field and $X$ be a vector space over $(K,|\cdot|)$. A function $p:X\to \mathbb{R}$ is a norm iff for each $a,b\in X$ and each $k\in K$, it satisfies:
In the case of a finite field, the valuation $|\cdot|$ must be the trivial one. In fact, if there is nonzero scalar $x\in K$ such that $|x|\neq1$, then $\{|x^n|:n\in\mathbb{Z}\}=\{|x|^n:n\in\mathbb{Z}\}$ is an infinite subset of $K$, which is a contradiction.
A comprehensive starting point to read about normed spaces in this context is the book: Non-Archimedean Functional Analysis - [A.C.M. van Rooij] - Dekker New York (1978).
For more information on finite fields, I recommend the paper: Non-archimedean Banach spaces over trivially valued fields, Borrey, S., P-adic functional analysis, Editorial Universidad de Santiago, Chile, 17 - 31. (1994). There, the norm is assumed to satisfy the strong triangle inequality.
For the study of more advanced stuff, like locally convex spaces over valued fields I recommend the book: Locally Convex Spaces over non-Arquimedean Valued Fields - [C.Perez-Garcia,W.H.Schikhof] - Cambridge Studies in Advanced Mathematics (2010).