As the others show, the answer to your question is affirmative. However, I don't see what's the point of doing so, when the problem can actually be converted into an unconstrained least square problem.
Let $Q$ be a real orthogonal matrix that has $\frac{v}{\|v\|}$ as its last column. For instance, you may take $Q$ as a Householder matrix:
$$
Q = I - 2uu^T,\ u = \frac{v-\|v\|e_n}{\|v-\|v\|e_n\|},
\ e_n=(0,0,\ldots,0,1)^T.
$$
Then $Xv=0$ means that $XQe_n=0$, which in turn implies that $XQ$ is a matrix of the form $XQ = (\widetilde{X},0)$, where $\widetilde{X}\in\mathbb{R}^{n\times(n-1)}$. So, if you partition $Q^TB$ as $Q^TB = \pmatrix{\widetilde{B}\\ \ast}$, where $\widetilde{B}$ is $(n-1)\times m$, then your minimisation problem can be rewritten as the unconstrained least-square problem
$$\min_{\widetilde{X}\in\mathbb{R}^{n\times(n-1)}} \|A-\widetilde{X}\widetilde{B}\|_F.$$
And the least-norm solution is given by $\widetilde{X} = A\widetilde{B}^+$, where $\widetilde{B}^+$ denotes the Moore-Penrose pseudoinverse of $\widetilde{B}$. Once $\widetilde{X}$ is obtained, you can recover $X = (\widetilde{X},0)Q^T$.
I shall prove that the problems are in fact equivalent, in the sense that they both have the same optimal solution $x^*$. Let $P_1$ be the first optimization problem, and $P_2$ the second, and let $\alpha\succ 0$.
If $\tilde{x}$ is the optimal solution for $P_1$, then we know from the book that each $\tilde{x}_i$ satisfies the relation
$$
\tilde{x}_i = \max\{0,1/\tilde{\nu} - \alpha_i\},
$$
where $\tilde{\nu}$ is the optimal dual variable associated with the equality constraint.
Now, write $P_2$ equivalently as
$$\begin{split}
\text{minimize} &\quad t\\
P_2:\quad\text{subject to}&\quad t\geq -\log(\alpha_i + x_i),\quad i=1 ,\ldots,n\\
&\quad x\succeq0,\quad \mathbf{1}^\top x=1.
\end{split}.$$
Then, the Lagrangian is
$$L(t,x,\mu,\lambda,\nu) = t - \sum_{i=1}^n \mu_i(\log(\alpha_i + x_i) + t) - \lambda^\top x + \nu(\mathbf{1}^\top x - 1) =\\t(1 - \mathbf{1}^\top\mu) - \sum_{i=1}^n\mu_i\log(\alpha_i + x_i) - \lambda^\top x + \nu(\mathbf{1}^\top x - 1).$$
Using this, you can write the KKT conditions of $P_2$ for the primal and dual optimal variables $t^*,x^*,\mu^*,\lambda^*$ and $\nu^*$ as
$$x^*\succeq 0,\quad \mathbf{1}^\top x^* = 1,\quad \lambda^*\succeq0,~\mu^*\succeq 0,\quad \lambda_i^*x_i^*=0,\quad t^* + \log(\alpha_i + x_i^*) \geq 0,\quad \mu_i^*(t^* + \log(\alpha_i + x^*_i)) = 0,\quad i=1,\ldots,n,$$
and
$$\partial/\partial x_i \left[ L(t^*,x^*,\mu^*,\lambda^*,\nu^*)\right]= -\mu_i^*/(\alpha_i^* + x_i^*) - \lambda_i^* + \nu ^* = 0,~i=1,\ldots,n,\quad \mathbf{1}^\top\mu^* = 1.$$
The last equality is necessary, otherwise the dual becomes unbounded in $t^*$. Eliminating $\lambda^*$ yields:
$$x^*\succeq 0,\quad \mathbf{1}^\top x^* = 1,\quad x_i^*(\nu^* - \mu_i^*/(\alpha_i + x_i^*))=0,\quad \nu^*\geq \mu_i^*/(\alpha_i + x_i^*),\quad i=1\ldots,n,$$
and
$$\mu^*\succeq 0,\quad \mathbf{1}^\top\mu^* = 1,\quad t^* + \log(\alpha_i + x_i^*) \geq 0,\quad \mu_i^*(t^* + \log(\alpha_i + x^*_i)) = 0,\quad i=1,\ldots,n.$$
These are in fact very similar conditions to those of $P_1$. Now, let $M = \{i\in \{1,\ldots,n\} ~|~ \mu_i^* > 0\}$. This is a nonempty set, because $\mathbf{1}^\top \mu^* = 1$. For all $i\in M$, you can thus deduce (in a similar way to the book) that in fact
$$x_i^* = \max\{0, \mu_i^*/\nu^* - \alpha_i\}.$$
Furthermore, we have that $t^* = -\log(\alpha_i + x_i^*)$.
Now, let $M' = \{1,\ldots,n\}\setminus M$. In this case, we have that for all $i\in M'$:
$$
x_i^*\nu^* = 0,\quad \nu^* \geq 0,\quad t^* + \log(\alpha_i + x_i^*) \geq 0.
$$
We can't have $\nu^* = 0$, otherwise it would imply that $0 < \mu_j^*/(a_j + x_j^*) \leq 0$, for all $j\in M$, which is impossible. So, the only option is that $x_i^* = 0$ for all $i\in M'$.
Thus, the problem reduces to solving the equation:
$$
\sum_{i\in M} x_i^* = \sum_{i\in M}\max\{0,\mu_i^*/\nu^* - \alpha_i\} = 1.
$$
However, the choice of $\mu$ can be arbitrary, as long as it satisfies the KKT conditions. Let $K = \{i\in\{1,\ldots,n\}~|~ \tilde{x}_i > 0\}$, and suppose we set the multipliers to
$$
\mu_i^* = \begin{cases}
1/\lvert K \rvert & i \in K\\
0 &\text{otherwise,}
\end{cases}
$$
then $\mathbf{1}^\top \mu^* = 1, \mu^*\succeq 0$. Clearly, we also have $M = K$, which implies $x_i^* = \tilde{x}_i=0$ for all $i\in M'$. Now, setting $\nu^* = \tilde{\nu}/\lvert M \rvert > 0$, then this implies $x_i^* = \tilde{x}_i$ for all $i\in M$, and we get exactly the same solution as $P_1$. And since $\tilde{x}_i + \alpha_i = \tilde{x}_j + \alpha_j$ for all $i,j\in M, i\neq j$, then this implies
$$
\log(x_i^* + \alpha_i) = \log(x_j^* + \alpha_j),
$$
thus $t^* = -\log(x_i^* + \alpha_i)$ for all $i\in M$, so all of the KKT conditions are satisfied.
Therefore, solving $P_2$ is equivalent to solving $P_1$. The optimal value of $P_2$ is $-\log(\alpha_i + \tilde{x}_i)$, for any $\tilde{x}_i > 0$.
Best Answer
Since $\Vert x\Vert_1\leq1$, this means $\sum_j\vert x_j\vert\leq 1$. Now let $x_0=-\mathrm{sign}(c_{i})e_i$ as in your question. For an arbitrary $x$ we then have: $$ |c^Tx|=\left\vert\sum_{j} c_j x_j\right\vert\leq\sum_j|c_j||x_j|\leq|c_i|\sum_j|x_j|=|c_i|\Vert x\Vert_1\leq|c_i| $$ Because $c^Tx_0=-|c_i|$, we have that this is indeed the best possible.