I'm wondering what's the matter with this property on oriented graphs :
$$ T= (V(T),E(T)) \text{ has prop } S_k \text{ if for any k vertices } v_1, \cdots, v_k \in V(T) \\ \text{ there exist a vertex } u \in V(T) \text{ such that } uv_1, uv_2, \cdots, uv_k \in E(T) $$
I think that $S_1$ is the necessary and sufficient condition for having a connected graph (is this even true? I don't know if we consider the direction of the edges), but what about this property for any k ? what is it saying?
$S_k$ says that every $k$ vertices have a common (direct) ancestor (predecessor may be the better, or more familiar term).
This does not say that the graph is connected for $k=1$.
For $k=1$ (and $T$ finite!) it does say that there is a directed cycle. The disjoint union of two directed cycles however satisfies $S_1$ but is not connected.
For $k>1$ and $T$ finite it does imply that the underlying simple graph is connected, but $T$ still does not need to be strongly connected.
For infinite graphs all bets are off: there need not even be cycles.
EDIT: above is for general digraphs (overlooking the fact that the question title clearly states tournaments).
In a tournament you can show that property $S_k$ implies minimum in-degree $k$ (provided $k$ is smaller than the size of the tournament). You can prove this as follows:
Suppose a vertex $v$ has in-degree $l<k$. Let $v_1u,\ldots,v_lu$ be the $l$ in-edges of $u$. Let $v_{l+1},\ldots,v_{k-1}$ other vertices (possibly zero if $l=k-1$), different from $u$, and all $v_i$ different from each other. By definition $u,v_1,v_2,\ldots,v_{k-1}$ have a common predecessor $w$ that must be different from all $v_i$ and therefore is another in-edge for $u$. Contradiction.