Yes. Let $L := \mathbf C \cap \partial B_1$ be the link of $\mathbf C$. Since $\mathbf C$ meets $\partial B_1$ transversely and is smooth near $\partial B_1$, $L$ can be viewed as a closed submanifold of $\mathbf S^n$ of dimension $n - 1$. But $H^{n - 1}(\mathbf S^n) = 0$, so there is an open subset $U \subset \mathbf S^n$ which is bounded by $L$. Then let $f$ be the indicator function of $$V := \left\{x \in \mathbf R^{n + 1}: \frac{x}{|x|} \in U\right\}.$$ The point is that $\mathbf C$ is an area-minimizing hypersurface which bounds the open set $V$; under these hypotheses we say that $V$ is a set of least perimeter (Bombieri and friends say that $V$ has an oriented boundary of least area).

It remains to show that if $f$ is the indicator function of a set of least perimeter then $f$ has least gradient. Suppose that it does not. Then there is a function $g$ such that $g - f$ has compact support in some ball $B$, but $$\int_B |dg| < \int_B |df|.$$ By the coarea formula, there exists $t \in \mathbf R$ such that $\partial \{g > t\}$ has less area than $\partial \{f > t\}$. Since $\partial \{f > t\}$ is empty for $t > 0$ we can take $t = 0$. But there exists a collar neighborhood $B' \subset B$ of $\partial B$ such that $f = g$ on $B'$. Therefore $$\partial \{g > 0\} \cap B' = \partial \{f > 0\} \cap B'.$$ But $\{f > 0\}$ has least perimeter so this is a contradiction.

Yes. Let $L := \mathbf C \cap \partial B_1$ be the link of $\mathbf C$. Since $\mathbf C$ meets $\partial B_1$ transversely and is smooth near $\partial B_1$, $L$ can be viewed as a closed submanifold of $\mathbf S^n$ of dimension $n - 1$. But $H^{n - 1}(\mathbf S^n) = 0$, so there is an open subset $U \subset \mathbf S^n$ which is bounded by $L$. Then let $f$ be the indicator function of $$V := \left\{x \in \mathbf R^{n + 1}: \frac{x}{|x|} \in U\right\}.$$ The point is that $\mathbf C$ is an area-minimizing hypersurface which bounds the open set $V$; under these hypotheses we say that $V$ is a set of least perimeter (Bombieri and friends say that $V$ has an oriented boundary of least area).

It remains to show that if $f$ is the indicator function of a set of least perimeter then $f$ has least gradient. Suppose that it does not. Then there is a function $g$ such that $g - f$ has compact support in some ball $B$, but $$\int_B |dg| < \int_B |df|.$$ By the coarea formula, there exists $t \in \mathbf R$ such that $\partial \{g > t\}$ has less area than $\partial \{f > t\}$. Since $\partial \{f > t\}$ is empty for $t > 0$ we can take $t = 0$. But there exists a collar neighborhood $B' \subset B$ of $\partial B$ such that $f = g$ on $B'$. Therefore $$\partial \{g > 0\} \cap B' = \partial \{f > 0\} \cap B'.$$ But $\{f > 0\}$ has least perimeter so this is a contradiction.

EDIT: Your typical function of least gradient has Cantor and jump parts, so usually "level set" in this context means "measure-theoretic boundary of a superlevel set". However, for an area-minimizing cone it can actually be arranged that the function of least gradient is absolutely continuous, in which case "measure-theoretic boundary of a superlevel set" really is equivalent to "level set" in the stronger sense that you're asking for.

To see this, recall that we can foliate the connected components of $\mathbf R^{n + 1} \setminus \mathbf C$ by area-minimizing hypersurfaces. This was proven by Hardt and Simon, but see also Lohkamp's "Minimal smoothings of area minimizing cones". Let $\mathbf C_s$, $s > 0$, be the hypersurfaces on one side of $\mathbf C$, and $\mathbf C_s$, $s < 0$, be the hypersurfaces on the other side, oriented so that $s \to 0$ near $\mathbf C$. Then let $f(x) = s$ whenever $x \in \mathbf C_s$, and $f(x) = 0$ on $\mathbf C$. The same argument with the coarea formula as above shows that this $f$ has least gradient.

In fact, it is very likely that your question is equivalent to the question of whether one can find a lamination $\lambda$ of area-minimizing hypersurfaces in $\mathbf R^{n + 1} \setminus \{0\}$ containing $\mathbf C \setminus \{0\}$ such that a sequence of leaves converges to $\mathbf C \setminus \{0\}$. The reason is that in dimension $\leq 7$, functions of least gradient are completely characterized by the fact that their level sets form a (Lipschitz, bounded curvature) lamination of area-minimizing hypersurfaces, as I showed in my recent preprint "Minimal laminations and level sets of $1$-harmonic functions", and it is natural to conjecture that the same result holds away from the singular set in any dimension (my argument does not prove that, anyways). For your problem we can of course take the Hardt-Simon foliation $\lambda$. But for more general area-minimizing hypersurfaces I suspect that this is not true.