For a given $\kappa > \omega$, define the game $d(\kappa)$ that runs for $\omega$ stages as follows: At stage $n$, player I chooses a sequence of elements of $\omega$, $g_n$ of length $\kappa$, and player II picks a natural number $f(n)$. Define $h_{\alpha} (n) = g_n (\alpha)$.
Player I wins the game if exists $\alpha \in \kappa$ such that $h_\alpha$ dominates $f$.
Let $\mathfrak{sd} = \min \left\lbrace \kappa : \text{player I has a winning strategy in } d(\kappa) \right\rbrace$.
$\mathfrak{sd}$ is uncountable by a diagonal argument and $\leq \mathfrak{d}$ since player I can play a dominating set.
Is $\mathfrak{sd} = \omega_1$?
Notice that player I has a winning strategy for the game $d(\kappa)$ iff there exists $\langle s_i : i < \kappa \rangle$ such that the following hold.
(1) Each $s_i: \omega^{\omega} \to \omega^{\omega}$.
(2) For every $n$, knowing $n$ bits of input gives you $n+1$ bits of output - i.e., for every $x, y \in \omega^{\omega}$, if $x \upharpoonright n = y \upharpoonright n$ then $s_i(x) \upharpoonright (n+1) = s_i(y) \upharpoonright (n+1)$.
(3) For every $x \in \omega^{\omega}$ there is some $i < \kappa$ such that $s_i(x)$ dominates $x$.
Now it is easy to check that for any $s_i$ as above, $\{x \in \omega^{\omega} : (\exists^{\infty} n)(s_i(n) = x(n))\}$ is comeager. Hence if $\kappa <$ cov(Meager), then player I does not have a winning strategy on $d(\kappa)$. In particular, $\mathfrak{sd} > \omega_1$ is consistent (add $\omega_2$ Cohen reals).
Although this answers your question, it is now natural to ask: Can $\mathfrak{sd} < \mathfrak{d}$?