Not being able to sleep due to jet lag, I had a chance to think about your problem a bit. The answer is elementary and simple, but I will say  more than is necessary.

First, people interested in quantitative information about $\ell^n_p$ have learned that it is usually better to to work with $L_p^n$ rather than $\ell^n_p$.  This simplifies formulas and sometimes even suggests better ways to view problems.  Denote the $L_p^n$ norm by $|\cdot|_p$, so $\|\cdot \|_p = n^{1/p}|\cdot |_p$.  So we are viewing $L_p^n$ as being $L_p(\mu_n)$, where $\mu_n$ is the uniform probability measure on $\{1,\dots,n\}$, and the $|\cdot |_p$ norms are increasing functions of $p$.  Also, since you are interested in norms on the left side of $2$, I will use $p$ (respectively, $q$) for your $p_*$ (respectively, $q_*$) and use a $*$ superscript rather than subscript for the dual norms.  

With this normalization, the answer to your problem is essentially independent of $n$, and you can formulate a version of your problem for $L_p : = L_p(0,1)$. I’ll also denote the norm on $L_p$ by $|\cdot |_p$.  
Coming back to $L_p^n$ from $L_p$   introduces a constant factor of at most $2$.


Problem: Fix a positive integer $T$ and let $1 \le p \le q \le 2$.  Compute the sup over all norm one operators $S: \ell_1^T \to L_q$ of
$$
\alpha_{p,q}(S) := \inf \{|Sx|_p : \|x\|_{\ell_1^T} = 1\}.
$$

Denote this supremum by $\alpha_{p,q}$.  So, as you (in effect) said at the beginning of your post, $\alpha_{q,q}=T^{-1/q^*}$. That for a norm one $S$ we have $\alpha_{q,q}(S) \le T^{-1/q^*}$ is immediate from the fact that $L_q$ has type $q$ with constant one (see e.g. the book of Albiac and Kalton).  The other inequality comes from considering an operator $S$ that maps the unit vector basis for $\ell_1^T$ to disjoint norm one functions in $L_q$.

Since $| \cdot |_r$ is an increasing function of $r$, it is clear that $\alpha_{p,q} \le \alpha_{q,q} $ for $1\le p \le q \le 2$. It remains to show that $\alpha_{p,q}  \ge T^{-1/q^*}$. To do this, take $T$ disjoint subsets $A_1,\dots,A_T$ of $(0,1)$ each having measure $T^{-1}$ and consider   
$S:\ell_1^T \to L_q$ defined by $Se_j := T^{-1/q} 1_{A_j}$; $1 \le j \le T$.

Coming back to $L_p^n$, we cannot quite do that last step if $T$ does not divide $n$, but you can choose the $T$ disjoint subsets of $\{1,\dots,n\}$ so that each set has measure between $(2T)^{-1} $ and $T^{-1}$.