It's true in $\Bbb R^n$, in fact you actually get an orthonormal basis. It's false for every ordered field other than $\Bbb R$.
Suppose first that $P\subset\Bbb R^n$ is the (strictly) positive cone of what I like to call a "linear linear order". A convex set with empty interior must be contained in a proper affine subspace, so $P$ has nonempty interior; say $v$ is an interior point of $P$.
Since $\overline{\Bbb R^n\setminus P}$ is closed and convex, a standard result shows that there exists (a unique) $p\in \overline{\Bbb R^n\setminus P}$ with $$||v-p||=\min_{q\in\overline{\Bbb R^n\setminus P}}||v-q||.$$ It follows that $$(x-p)\cdot(v-p)\le0\quad(x\in\Bbb R^n\setminus P).$$(Draw a picture; note that those two vectors meet at an angle greater than or equal to $\pi/2$. Or assume that $(x-p)\cdot(v-p)>0$, let $q_t=tx+(1-t)p=p+t(x-p)$ for $t\in(0,1)$ and show that $||q_t-p||<||v-p||$ if $t$ is small enough).
Since $x\in\overline{\Bbb R^n\setminus P}$ and $r>0$ imply $rx\in\overline{\Bbb R^n\setminus P}$ this shows that $$rx\cdot(v-p)\le p\cdot(v-p)\quad(x\in\overline{\Bbb R^n\setminus P},r>0).$$Letting $r$ tend to infinity shows that $$x\cdot(v-p)\le0\quad(x\in-P),$$hence $$x\cdot(v-p)\ge0\quad(x\in P).$$
And now we're on our way. Let $v_1=(v-p)/||v-p||$. If $x\cdot v_1>0$ then $x>0$, if $x\cdot v_1<0$ then $x<0$, while if $x\cdot v=0$ we don't know yet. But induction on the dimension shows that there is an orthonomal basis $v_2,\dots,v_n$ for $v_1^\perp$ such that our order restricted to $v_1^\perp$ is the lexicographic order determined by $v_2,\dots,v_n$. Hence $<$ is the lexicographic order determined by $v_1,\dots,v_n$. QED.
As an aid to seeing what's going on in the construction for $\Bbb K\ne\Bbb R$, note that it's inspired by the following simple example for $\Bbb K=\Bbb Q$: If $$P=\{x\in\Bbb Q^2:x_1+\sqrt 2x_2>0\}$$then $P$ is the positive cone of an order which is not the lexicographic order determined by a basis. We can do essentially the same thing in any field other than $\Bbb R$, except we have to concoct a Dedekind cut to use in place of $-1/\sqrt2$.
Suppose then that $\Bbb K$ is an ordered field not isomorphic to $\Bbb R$. Then $\Bbb K$ is not order-complete, so it's not hard to show that there exist nonempty sets $A$ and $B$ with $\Bbb K=A\cup B$, such that every element of $A$ is strictly smaller than every element of $B$, and such that $A$ has no largest element while $B$ has no smallest element.
Let $P$ be the set of $x\in\Bbb K^2$ such that there exist $\alpha\in A$ and $\beta\in B$ with $$x_1+\gamma x_2>0\quad(\gamma\in(\alpha,\beta)).$$If $x\in P$ and also $-x\in P$ then there exists $\gamma$ with $x_1+\gamma x_2>0$ and also $-(x_1+\gamma x_2)>0$, contradiction.
Suppose on the other hand that $x\notin P$ and also $-x\notin P$. If $x_2=0$ then $x\notin P$ shows that $x_1\le0$, while $-x\notin P$ shows that $-x_1\le0$; hence $x=0$. Suppose $x_2\ne0$. Replacing $x$ by $-x$ if necessary, wlog $x_2>0$. Now $x\notin P$ shows that $-x_1/x_2>\alpha$ for every $\alpha\in A$, while $-x\notin P$ shows that $-x_1/x_2<\beta$ for every $\beta \in B$. In particular $-x_1/x_2\notin A\cup B=\Bbb K$, contradiction.
So the three sets $P$, $\{0\}$, $-P$ form a partition of $\Bbb K^2$. Since $P$ is closed under addition and closed under multiplication by positive scalars it follows that $P$ is the positive cone of some order $<$ on $\Bbb K^2$.
And $<$ is not the lexicographic order determined by some basis. In fact there does not exist a non-zero linear functional $\Lambda$ such that $\Lambda x>0$ implies $x\in P$. This seems clear; a proof by eight or sixteen special cases depending on the signs of various things is trivial, although I never did quite work out the details. Finally found a nicer proof:
For $\gamma\in\Bbb K$ let $$E_\gamma=\{x\in\Bbb K^2:x_1+\gamma x_2>0\}.$$Then $E_\gamma$ is open. If $\gamma\in[\gamma_1,\gamma_2]$ then $\gamma$ is a convex combination of $\gamma_1$ and $\gamma_2$; this shows that $$ E_{\gamma_1}\cap E_{\gamma_2}\subset E_\gamma\quad(\gamma\in[\gamma_1,\gamma_2])).$$And that shows that $P$ is open. Indeed, the definition of $P$ is $$P=\bigcup_{\alpha\in A,\beta\in B}\bigcap_{\gamma\in(\alpha,\beta)}E_\gamma,$$but that inclusion shows that in fact $$P=\bigcup_{\gamma_1\in A,\gamma_2\in B}E_{\gamma_1}\cap E_{\gamma_2}.$$
So no point of $P$ is a boundary point of $P$. And $-P$ is open, so no point of $-P$ is a boundary point of $P$. So $$\partial P=\{0\}.$$
Suppose now that $\Lambda$ is a nonzero linear functional such that $\Lambda x>0$ implies $x\in P$. Then $\Lambda x\ge0$ implies $x\in\overline P$. And so $\Lambda x\le 0$ implies $x\in\overline{-P}$.
So $\Lambda x=0$ implies $x\in\partial P=\{0\}$. There is no such $\Lambda$.
