Surprisingly (at least to me) the answer is no when $n\ge 3$. This was proved by Yves Benoist and me after I mentioned the problem in a talk at MSRI and Yves came up with a great idea.
It is enough to show that there is no uniform bound when $n=3$.
Here is an elementary argument that involves little computation. I don't pay attention to getting exact constants.
For small positive $\epsilon$ define
\begin{equation}
x_1 = e_1 \quad \quad
x_2 = e_1 + \epsilon e_2 \quad \quad
x_3 = e_1 + \epsilon e_3
\end{equation}`
For certain distinct non zero real $\lambda_i$, which will depend on $\epsilon$, let $T\in M_n(\mathbb{R})$ be defined by $Tx_i = \lambda_i x_i$. The dual basis to $(x_i)$ is defined by
\begin{equation}
f_1=e_1 - {1\over\epsilon} (e_2 + e_3) \quad \quad
f_2={1\over\epsilon} e_2 \quad \quad
f_3={1\over\epsilon} e_3
\end{equation}
So the transpose and adjoint of $T$ is defined by $T^*f_i = \lambda_i f_i$. If $S$ implements a similarity between $T$ and $T^*$, it is more or less clear that $\|S\|\cdot \|S^{-1}\| \to \infty$ as $\epsilon \to 0$ uniformly over all permissible choices of $\lambda_i$. (For me the easy way to see this is to semi-normalize the $f_i$ as
\begin{equation}
\tilde{f}_1=\epsilon e_1 - (e_2 + e_3) \quad \quad
\tilde{f}_2= e_2 \quad \quad
\tilde{f}_3= e_3
\end{equation}
The similarity $S$ must be given by $Sx_i = a_i \tilde{f}_i$ for some non zero $a_i$ since the $\lambda_i$ are distinct, and if $\|S\| \vee \|S^{-1}\| \le (1/3) C$, then all $|a_i|$ and all $1/|a_i|$ are less than $C$. But $\|x_2-x_3\|=\sqrt{2}\epsilon$ and
$\|a_2 \tilde{f}_2 - a_3 \tilde{f}_3 \| =\sqrt{|a_2|^2+|a_3|^2} > \sqrt{2} /C$, which forces
$\|S\| >1/(C\epsilon)$.)
Next a trivial but (it seems) important point. For fixed $\epsilon$, you can choose the $\lambda_i$ close enough to one so that $T$ is as close to the identity as you want. So we can choose $\lambda_i$ so that $\|T\|=1$ and $\|T^{-1}\|< 1+\epsilon$; denote such a $T$ by $T_\epsilon$.
Write $T_\epsilon = H_\epsilon K_\epsilon$ with $H_\epsilon$, $K_\epsilon$ (complex) Hermitian and $\|H_\epsilon\|=1$. We want to see that $\|K_\epsilon\| \to \infty$ as $\epsilon \to 0$. Notice that $H_\epsilon$ and $K_\epsilon$ are non singular since no $\lambda_i$ is zero. So we have
$H_\epsilon^{-1}T_\epsilon H_\epsilon = K_\epsilon H_\epsilon = T_\epsilon^*$ and hence, by the first part of the proof, $\|H_\epsilon^{-1}\| \to \infty$ as $\epsilon \to 0$. But $H_\epsilon^{-1} = K_\epsilon T_\epsilon^{-1}$, so
$$\|H_\epsilon^{-1}\| \le \|K_\epsilon\| \|T_\epsilon^{-1}\| \le (1+\epsilon) \|K_\epsilon\|.
$$
Set $$H = \left( \begin{matrix} 1 & 1 \\ 1 & -1 \end{matrix} \right).$$
Then, the matrix $H_k := H^{\otimes k}$ is a $2^k \times 2^k$-matrix with entries $\pm 1$ (with respect to the natural basis). Set $n:=2^k$. It is easy to see that $H_k^2=n$.
It follows that the sum of absolute values of eigenvalues of $H_k$ is $n^{3/2}$. So, no upper bound better than this can be expected.
Best Answer
As noted in the comments, the quantity you want is $\sigma_{\min}(V)^{-2}$, the inverse square of the minimum singular value of $V$.
Unfortunately you can't get any meaningful bound from below for matrices with unit column norms, as they can still be arbitrarily close to singular. For instance, $$ \begin{bmatrix} 1 & \cos \alpha \\ 0 & \sin \alpha \\ 0 & 0 \end{bmatrix} $$ has a singular value that must tend to $0$ when $\alpha \to 0$.