In some degenerate cases there may be no such a one point (for instance, if all the lines are parallel). However there's a single solution in the general case.
I assume you're trying to solve a more general problem where all the lines are not required to intersect exactly (otherwise there's a much simpler solution than the least squares).
Derivation:
You say the every line is represented by two points. Let's rather work in the convention where a line is represented by one point and a direction vector, which is just a vector subtraction of those two points. That is, instead of describing a line by points $\mathbf{a}$ and $\mathbf{b}$ we'll describe it by a point $\mathbf{a}$ and a vector $\mathbf{d}$ whereas $\mathbf{d}=\mathbf{b}-\mathbf{a}$.
Our point (which we're trying to find) is $\mathbf{c}$.
The distance of this point to the line is:
$H=\frac{\|(\mathbf{c}-\mathbf{a})\times\mathbf{d}\|}{\|\mathbf{d}\|}$
Using identity $(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{a}\times\mathbf{b})=\|\mathbf{a}\|^2\|\mathbf{b}\|^2-(\mathbf{a}\cdot\mathbf{b})^2$
we have:
$H^2=\frac{\|\mathbf{c}-\mathbf{a}\|^2\|\mathbf{d}\|^2-\|(\mathbf{c}-\mathbf{a})\cdot\mathbf{d}\|^2 }{\|\mathbf{d}\|^2}$
$H^2 = \|\mathbf{c}-\mathbf{a}\|^2-\frac{\|(\mathbf{c}-\mathbf{a})\cdot\mathbf{d}\|^2 }{\|\mathbf{d}\|^2}$
The square sum of the distances of the point $\mathbf{c}$ to all the lines is just the sum of the above expressions for all the lines. The problem is to minimize this sum. This sum depends on a variable $\mathbf{c}$ (which is actually 3 variables, the components of $\mathbf{c}$). This is a standard least squares problem, which generally has a single solution (unless there's a degeneracy).
Solving the least squares for this specific case.
Since we want find such a $\mathbf{c}$ that minimizes this sum, its derivative with regard to $\mathbf{c}$ should be zero. In other words:
$\frac{d(H^2)}{d\mathbf{c}}=2(\mathbf{c}-\mathbf{a})-2\mathbf{d}\frac{(\mathbf{c}-\mathbf{a})\cdot\mathbf{d}}{\|\mathbf{d}\|^2}$
$0=\sum_{i=0}^m{\mathbf{c}-\mathbf{a}^{(i)}-\mathbf{d}^{(i)}\frac{(\mathbf{c}-\mathbf{a}^{(i)})\cdot\mathbf{d}^{(i)}}{\|\mathbf{d}^{(i)}\|^2}}$
This gives 3 equations (since it's a vector equation) with 3 unknowns (components of $\mathbf{c}$).
This is a problem of finding a geometric median on a sphere. While the mean provides a minimizer for the sum of squared distances (in a Euclidean setting), the $L^1$ minimum is generally not expressible in such an explicit form. For three points, the Euclidean solution is known as a Fermat point, and for four coplanar points an explicit solution is known.
The usual approach is an iterative one, using a weighted least squares minimizer while adjusting the weights to obtain the $L^1$ minimum. The basic method is called the Weiszfeld algorithm, and because of the convexity of the distance function, a suitable variant on the sphere has been conjectured to converge except for countably many initial points (though as noted, the minimizing location is not necessarily unique).
Added: Let me point out a note on the three point case by K. Ghalieh and M. Hajja, The Fermat point of a spherical triangle in The Mathematical Gazette of Nov. 1996 (pp. 561-564). Although it is "behind a pay wall", you can nonetheless take advantage of JSTOR's free-of-charge program (registration required) to read the article online (see site for details), as I have done.
The authors show that when the sides of a spherical triangle $ABC$ are sufficiently short, each less than $\pi/2$ on a sphere of unit radius (implying that they are not all on the same great circle and that no two vertices are antipodal), then a Fermat point (minimizing the sum of spherical distances to three vertices) exists, is unique, and shares some defining properties with the case of a triangle in the Euclidean plane:
- If all three vertex angles are less than 120°, then the Fermat point $P$ is "inside" the spherical triangle (in the smaller region of the sphere bounded by the triangle) and the three sides subtend equal angles around $P$:
$$ \angle APB = \angle BPC = \angle CPA = 120° $$
- If one of the vertex angles is 120° or more, that vertex is the Fermat point.
Best Answer
I will switch to $x^1, \dots, x^n$ as the notation for $\mathcal X$, so that I can use subscripts for components of a vector: $x^i = (x^i_1, x^i_2, \dots, x^i_p)$.
Let's begin by understanding the function $$ f(x) = \sum_{i=1}^n \|x - x^i\|_2^2 = \sum_{i=1}^n \sum_{j=1}^p (x_j - x^i_j)^2. $$ Switching the order of summation and taking the $j^{\text{th}}$ term of the sum over $j$, we get $\sum_{i=1}^n (x_j - x^i_j)^2$. This is a quadratic equation in $x_j$ where the coefficient of $x_j^2$ is $n$, so it can be written as $n(x_j - h_j)^2 + k_j$ for some $h_j, k_j \in \mathbb R$. Summing these together again, we get $$ f(x) = \sum_{j=1}^p n(x_j - h_j)^2 + k_j = n \|x - h\|_2^2 + \sum_{j=1}^p k_j. $$ In other words, $f(x)$ is $n$ times the distance to some point $h$, plus a constant.
By the characterization of $\mu$ as the point minimizing $f(x)$, we know that actually $h = \mu$. This tells us what the additive constant must be: it is $f(\mu)$ (the sum of distance from $\mu$ to the points in $\mathcal X$). We conclude that $$ f(x) = n\|x - \mu\|_2^2 + f(\mu). $$ The point $x^i \in \mathcal X$ closest to all other points in $\mathcal X$ is the point $x^i$ for which $f(x^i)$ is the smallest. We see now that it must be the point for which $\|x^i - \mu\|_2$ is the smallest, therefore your equality holds.
Regarding the possibility of the same thing happening with other norms: the key thing that makes this work for $\ell_2$ is the fact that the sum of squared distance functions is a multiple of a distance function. (Plus a constant.) In the simplest case, for two points $y, z$, we have $$ \|x - y\|_2^2 + \|x - z\|_2^2 = 2\|x - \tfrac{y+z}{2}\|_2^2 + \frac12\|y-z\|_2^2. $$ This lets us rewrite the function $f(x)$ above in terms of distance from the mean.
It is easy to check that the same "coincidence" does not occur for other norms, so when we work with a different norm, the function $f(x)$ will have a more complicated shape: it will not solely depend on the distance from some center point.
When $\mathcal X$ is very large, adding a new point or two will not change the shape of $f(x)$ very much. So we can add two new points $x^{n+1}, x^{n+2}$ with the following properties:
Then the first argmin in the question will be $x^{n+1}$, and the second will be $x^{n+2}$; this outlines how to come up with a counterexample for other norms.