$\text{colim}(\mathcal{C} \hookrightarrow \mathcal{X}) = 1$ A full subcategory $\mathcal{C} \subset \mathcal{X}$ is dense if all $X \in \mathcal{X}$ is a colimit of object in $\mathcal{C}$. More precisely, $X = \text{colim}(\text{pr}:\mathcal{C}/X \rightarrow \mathcal{X})$.
In this conditions, $L := \text{colim}(\mathcal{C} \hookrightarrow \mathcal{X}) = 1$.
If $1$ (terminal object) exists, the cone form by $1$ and the unique arrows to it, is a universal cone.
If $L$ exists, I have an arrow $X \rightarrow L$ cause $X$ is a universal cone. But I can't see why is this arrow unique?
 A: I think the result does not hold. Here is a (minimal ?) counterexample : 
consider for $\mathcal X$ a category with three objects, $X,Y,L$ and four nonidentity maps : $f,g :X\to L, h: Y\to L, k:L\to L$ subject to the following conditions :
$f\neq g$, $k\circ f= g$, $k\circ k= id_L$ (and the following relations follow from this : $k\circ h = h, k\circ g = f$)
Then let $\mathcal C= $ the full subcategory on $X,Y$. I claim that it is dense. Indeed $X,Y$ are their own colimit, and I claim that $(L,f,h)$ is a coproduct of $X$ and $Y$. 
Proof : let $Z$ be an object with maps $q_0 : X\to Z, q_1: Y\to Z$. Then necessarily $Z=L$, $q_1= h$, and $q_0$ is either $f$ or $g$. 
If $q_0 = f$ then clearly $id_L :L\to L$ makes the appropriate diagram commute, and is unique in doing so (because $k$ doesn't !)
If $q_0=g$, then clearly $k:L\to L$ makes the appropriate diagram commute, and is unique in doing so (because $id_L$ doesn't).
It follows that $(L,f,h)$ is a coproduct indeed. 
Moreover, $\mathcal C$ is a discrete category on two objects, and so $\mathcal {C\to X}$ is just a coproduct diagram : its colimit is the same as a coproduct of $X,Y$, so its colimit is $L$.
However, $L$ is not final in $\mathcal X$. So the claim is false. 
Let me edit to make what I meant more precise and to actually follow Kevin Carlson's comment below : the claim that is false is under the assumption that "any object of $\mathcal X$ is a colimit of some objects in $\mathcal C$", which is a weaker claim than what followed "more precisely", which is the claim that $X\in \mathcal X$ is "canonically" a colimit over $\mathcal C$. 
If you have that stronger condition then the claim does hold. The essential part of the argument is proving that there is only one morphism $X\to L$ for $X\in \mathcal C$. 
Indeed suppose that this holds and let $X\in \mathcal C$. Write $X$ as a colimit $\mathrm{colim}_i C_i$ of objects in $\mathcal C$. Then $\hom(X,L) = \lim_i \hom(C_i,L) = \lim_i *$ by the assumption, so $\hom(X,L) = *$, which proves that $L$ is terminal. Note that in this part of the proof, we only require the weaker condition, so it's in the first part that the stronger condition will be needed (and indeed, that's how I looked for my counterexample)
For the first part, write $L$ as two colimits : one as the cocone $(L, (\mu_X)_{X\in\mathcal C})$  and the other as the cocone $(L,(f)_{f \in \mathcal C/L})$. Note that here I'm making an extra assumption which is implicit in your post but not fully precise there: the identification $L= \mathrm{colim}(\mathcal C/L\to\mathcal X)$ comes from the natural cocone on $L$. 
I don't know if that assumption can be removed, I don't seem to be getting anywhere without it. 
With that assumption, we automatically get from the universal properties involved a map $\kappa : L\to L$ such that $\kappa \circ f= \mu_X$ for all $f: X\to L\in \mathcal C/L$. 
In particular, $\kappa\circ \mu_X = \mu_X$ for all $X$, so that $\kappa= id_L$, so that $f= \mu_X$ for all $f:X\to L\in \mathcal C/L$. In other words, for $X\in \mathcal C, \hom(X,L) = \{\mu_X\}$, which is what we wanted. 
