How to find a countable complete theory $T$ such that $f_T(\kappa)=(\text{ded}(\kappa))^{\aleph_0}$ holds for every infinite cardinal $\kappa$?
Here $f_T(\kappa)$ is defined as $\sup\{|S_1(M)|\,|\,M\models T,\,|M|=\kappa\}$ and $\text{ded}(\kappa)$ is defined as $\sup\{|I|\,|\,I \text{ contains a dense subset of size }\kappa\}=\sup\{\lambda\,|\,\text{there is a linear order of size }\kappa\text{ with }\lambda\text{ many cuts}\}$.
Consider the language $\{<_n\mid n\in \omega\}$. Let $T$ be the theory which says that each $<_n$ is a linear order, and let $T^*$ be the model companion of $T$. More concretely: $T^*$ says that each $<_n$ is a dense linear order without endpoints, and for all $N\in \omega$, if $I_i$ is an interval in the order $<_i$ for all $i<N$, then $\bigcap_{i<N}I_i\neq \varnothing$.
Now $T^*$ has quantifier elimination, so a complete $1$-type over $M$ is determined by all instances of $x <_n m$ and $m<_n x$ and their negations, for $m\in M$ and $n\in \omega$. Every $1$-type over $M$ is either realized in $M$, or it specifies a cut in $(M,<_n)$ for all $n\in \omega$. Thus if $|M| = \kappa$, we have $|S_1(M)|\leq (\mathrm{ded}(\kappa))^{\aleph_0}$. So $$f_{T^*}(\kappa)\leq (\mathrm{ded}(\kappa))^{\aleph_0}.$$
To obtain equality here is slightly more complicated, because $\mathrm{ded}(\kappa)$ is defined to be the supremum of the number of dedekind cuts in linear orders of size $\kappa$, and this supremum might not be obtained.
Case 1: $\mathrm{ded}(\kappa)$ is a successor cardinal. Then the supremum is attained: there is a linear order $(L,<)$ of cardinality $\kappa$ with $\mathrm{ded}(\kappa)$-many dedekind cuts. Defining $M$ of cardinality $\kappa$ so that $(L,<)$ embeds in $(M,<_n)$ for all $n$, we have $|S_1(M)| = (\mathrm{ded}(\kappa))^{\aleph_0}$, so $f_{T^*}(\kappa)=(\mathrm{ded}(\kappa))^{\aleph_0}$.
Case 2: $\mathrm{ded}(\kappa)$ is a limit cardinal with $\mathrm{cf}(\mathrm{ded}(\kappa)) = \aleph_0$. In this case, the supremum is also attained. Write $\mathrm{ded}(\kappa) = \sup_{n\in \omega} \lambda_n$ with each $\lambda_n<\mathrm{ded}(\kappa)$. For each $\lambda_n$, pick a linear order $(L_n,<)$ of cardinality $\kappa$ with at least $\lambda_n$-many dedekind cuts. Now the union of the $L_n$ has cardinality $\kappa$ and has $\mathrm{ded}(\kappa)$-many dedekind cuts. Proceed as in Case 1. The same argument shows that the supremum is attained whenever $\mathrm{cf}(\mathrm{ded}(\kappa)) \leq \kappa$.
Case 3: $\mathrm{ded}(\kappa)$ is a limit cardinal with $\mathrm{cf}(\mathrm{ded}(\kappa)) > \aleph_0$. For each $\lambda<\mathrm{ded}(\kappa)$, we can find a linear order $(L_\lambda,<)$ of cardinality $\kappa$ with at least $\lambda$-many dedekind cuts. Defining $M_\lambda$ of cardinality $\kappa$ so that $(L_\lambda,<)$ embeds in $(M_\lambda,<_n)$ for all $n$, we have $|S_1(M_\lambda)| \geq \lambda^{\aleph_0}$. Then $$f_{T^*}(\kappa) = \sup\{|S_1(M)|\mid M\models T, |M| = \kappa\} \geq \sup_{\lambda<\mathrm{ded}(\kappa)} \lambda^{\aleph_0}.$$ But since $\mathrm{cf}(\mathrm{ded}(\kappa))>\aleph_0$, every function $\aleph_0\to \mathrm{ded}(\kappa)$ is bounded and hence can be considered as a function $\aleph_0\to \lambda$ for some $\lambda < \mathrm{ded}(\kappa)$, so $$\sup_{\lambda<\mathrm{ded}(\kappa)} \lambda^{\aleph_0} = (\mathrm{ded}(\kappa))^{\aleph_0}.$$
Because of this example, Keisler calls theories with $f_T(\kappa) = (\mathrm{ded}(\kappa))^{\aleph_0}$ "multiply ordered". For more examples, see Keisler's expository paper on $f_T$: Six classes of theories.