This answer is to the clarified version of the question in your comment. We can show that no, we cannot write $C_1$ as a function of $\|x_1\|, \ldots, \|x_n\|$.
Take, for example, the space $X = \Bbb{R}^2$ (with no norm just yet) and the basis $x_1 = (-1, 1)$ and $x_2 = (1, 1)$. We are going to define a sequence of norms $(\|\cdot\|_k)_{k=1}^\infty$ on $X$ that each map $x_1$ and $x_2$ to the same number, but for which the lower bound $C_1^{(k)}$ tends to $0$ as $k \to \infty$. So, even though all the norms evaluate the basis the same way, there is no universal lower bound $C_1 > 0$ that will satisfy the desired inequality.
The intuitive idea here is that we can build norms from convex sets. In particular, a set $B$ is the closed unit ball of a norm on $\Bbb{R}^n$ if and only if $B$ is convex, (linearly) bounded, symmetric (i.e. if $x \in B$, then $-x \in B$), closed and has non-empty interior (with respect to the Euclidean topology, shared by all norms).
The larger $B$ is, the smaller we are forced to make $C_1$. If we keep $x_1, x_2$ on the boundary of $B$, we keep $\|x_1\| = \|x_2\| = 1$. So, we simply define norms based on increasingly large $B$, that keep $x_1, x_2$ on the boundary. I picture a series of ellipses, with the $y$-axis aligning with its major axis, and the $x$-axis aligning with its minor axis. I see that the major axis continues to grow (and the minor axis will accordingly have to shrink to fit).
Norms whose balls are ellipses of this form take the form
$$\|(x, y)\|^2 = \frac{x^2}{a^2} + \frac{y^2}{b^2}$$
for constants $a$ and $b$. We want $b > a$, so that the $y$ axis is the major axis. Indeed, we are looking for $b$ to increase to $\infty$ in our sequence of norms. We also want $\|(\pm1, 1)\| = 1$, so
$$1 = \frac{1}{a^2} + \frac{1}{b^2} \implies a^2 = \frac{b^2}{b^2 - 1}.$$
With this in mind, we shall choose $b = \sqrt{n + 1}$, giving us:
$$\|(x, y)\|_n^2 = \frac{n}{n+1}x^2 + \frac{1}{n + 1}y^2.$$
It's not difficult to see that this is a norm, and in fact it's derived from an inner product:
$$\langle (a, b), (c, d) \rangle = \frac{n}{n+1}ac + \frac{1}{n+1}bd.$$
Clearly $\|x_1\|_n = \|x_2\|_n = 1$ for all $n$. We just need to show that any choice of lower bound $C^{(n)}_1$ tends to $0$ as $n \to \infty$.
Consider, in particular, $(0, 2) = x_1 + x_2$. If we have $C_1 = C_1^{(n)}$ and $C_2 = C_2^{(n)}$ as you wrote in your question, then
$$0 \le 2C_1^{(n)} \le \|(0, 2)\|_n^2 \le 2C^{(n)}_2.$$
But,
$$\|(0, 2)\|_n^2 = \frac{n}{n+1} \cdot 0 + \frac{1}{n+1} \cdot 2 \to 0$$
as $n \to \infty$. By squeeze theorem, we must have any choice of $C_1^{(n)} \to 0$ as $n \to \infty$.
Best Answer
Consider $X=\ell^\infty$ with $\sup$ norm. For $0<\varepsilon <{1\over n-1}$ and $1\le k\le n$ let $$x_k=(1-\varepsilon){\mathbf 1}+\varepsilon\delta_k,$$ where $\delta_k$ denotes the sequence with entry $1$ on the $k$-th coordinate and $0$ otherwise, while ${\mathbf 1}$ denotes the constant sequence with all coordinates equal $1.$ Then $\|x_k\|=1,$ $\|x_k-x_l\|=\varepsilon, $ for $k\neq l,$ and $$\|x_1+x_2+\ldots +x_n\|=n-(n-1)\varepsilon $$ By the triangle inequality for any normed space we have $$\|x_2+x_2+\ldots +x_n\|\ge n-(n-1)\varepsilon ,\quad \|x_k\|=1$$ Hence the lower estimate is the worst possible.