Your analysis is wrong: $SU_2$ is the universal cover of $SO_3(\mathbb R)$ but (1) no such relation exists for general rank (note that already the indices are not the same between the two groups), and more importantly (2) the types $D_n$ and $B_n$ refer to complex groups $SO_{2n}(\mathbb C)$ and $SO_{2n+1}(\mathbb C)$ which are not in any special relation to any real group $SU_n$, or the corresponding complex group $SL_n(\mathbb C)$, at all. So types $B,D$ are really different from type $A$ (as are types $C$ of course), and one does not get any other classical types for free by doing type $A$. In fact, especially for questions of tableaux, the other classical types are considerbly harder to treat than type $A$.
To answer your question, there do exist generalisations of tableaux for other classical types that capture some aspects of what Young tableaux do for type $A$. However, unfortunately the type of generalisation to use depends on the particular aspect of interest, and so there are about as many different generalisations as there are papers on these subjects. It is a great mess; in some cases there are (rather involved) bijections between certain types of tableaux, or other rather more subtle relations, but not in all cases, and I don't think there is even a good inventory of all the types of tableaux that have been invented.
I claim that $N:=N_{SU(4)}(Sp(2))$ is a maximal subgroup of $Sp(2)$, and that $N = Sp(2) \cup iI Sp(2)$.
To see this, consider the double covering $\pi:SU(4)\rightarrow SU(4)/\{\pm I\}\cong SO(6)$. Note that $-I\in Sp(2)\subseteq SU(4)$, so $\pi|_{Sp(2)}$ is the double covering $Sp(2)\rightarrow SO(5)$. Up to conjugacy, there is an essentially unique $SO(5)\subseteq SO(6)$, the usual block form.
So, instead of studying $Sp(2)\subseteq SU(4)$, we'll study $SO(5)\subseteq SO(6)$ and pull the information back via $\pi$.
Proposition: The only proper subgroup of $SO(6)$ which properly contains $SO(5)$ is $O(5) = \{ \operatorname{diag}(A,\det(A)):A\in O(5)\}$.
Proof: The isotropy action of $SO(5)$ on $S^5 = SO(6)/SO(5)$ is transitive on the unit sphere in $T_{I SO(5)} S^5$, so is, in particular, irreducible. This, then, implies that $SO(5)$ is maximal among connected groups: if $SO(5)\subseteq K\subseteq SO(6)$, then on the Lie algebra level, the adjoint action of $\mathfrak{so}(5)$ would preserve both $\mathfrak{k}$ and $\mathfrak{k}^\bot$, contradicting irreducibility. (Here, I'm using the fact that we can naturally identify the isotropy action of $H$ on $T_{I SO(5)} SO(6)/SO(5)$ with $\mathfrak{so}(5)^\bot\subseteq \mathfrak{so}(6)$.)
Since we now know that $SO(5)$ is maximal among connected subgroups of $SO(6)$, and the identity component of a Lie group is always a normal subgroup, it now follows that $N_{SO(6)}(SO(5))$ is a maximal subgroup of $SO(6)$.
Of course, $O(5)\subseteq N_{SO(6)}(SO(5))$, but why is the reverse inclusion true? Well, every matrix $B\in SO(5)$ fixes the basis vector $e_6$ of $\mathbb{R}^6$, and $\operatorname{span}\{e_6\}$ is the unique subspace of $\mathbb{R}^6$ fixed by all of $SO(5)$. A simple computation reveals that for any $C\in N_{SO(6)}(SO(5))$, that $CSO(5)C^{-1} = SO(5)$ fixes $Ce_6$. It follows that $Ce_6 \in \operatorname{span}\{e_6\}$. Moreover, since $C\in SO(6)$, we must in fact that $Ce_6 = \pm e_6$. In either case, the fact that $CC^t = I$ now implies that $C\in O(5)$. $\square$
Now, let's pull that information back to to better understand $N = N_{SU(4)}(Sp(2))$. It's not too hard to see that $\pi|_N:N\rightarrow N_{SO(6)}(SO(5)$ is a double covering, with $\pi$ mapping $Sp(2)$ and $iI Sp(2)$ to the two different components of $N_{SO(6)}(SO(5))$.
Now, let $g\in N$ be arbitrary. Since $\pi|_{Sp(2)\cup iI Sp(2)}$ is surjective onto $N_{SO(6)}(SO(5))$, there is an $h\in Sp(2)\cup iI Sp(2)$ with $\pi(g) = \pi(h)$. Then $gh^{-1}\in \ker \pi = \pm I$, so $g = \pm I h$. Since both $h, \pm I\in Sp(2)\cup iI Sp(2)$, it follows that $g\in Sp(2)\cup iI Sp(2)$ as well.
Best Answer
$\def\fg{\mathfrak{g}}\def\ft{\mathfrak{t}}\def\long{\text{long}}$This question was crossposted to MO
https://mathoverflow.net/questions/435457/normalizer-of-maximal-torus-is-maximal-in-compact-simple-group/435501#435501
Where Mikhail Borovoi gave a short conceptual proof
https://mathoverflow.net/a/435501/387190
that if all the roots have the same length (types $ A_n,D_n,E_6,E_7,E_8 $) then $ N(T) $ is maximal.
And David E Speyer gave a nice conceptual proof
https://mathoverflow.net/a/435474/387190
that if the roots are not all the same length (i.e. there are long roots and short roots i.e. types $ B_n,C_n,F_4,G_2 $) then $ N(T) $ is not maximal since it is contained in the proper closed subgroup of strictly larger dimension $$ G_{long}N(T)=N(G_{long}) $$ whose Lie algebra is spanned by the lie algebra of $ T $ together with the roots spaces for all the long roots.
In particular let me highlight this part of the answer from David E Speyer:
In type $B_n = SO(2n+1)$, the group $G_{\long}$ is $SO(2n)$ and $G_{\long} N(T) = O(2n)$.
In type $C_n = Sp(2n)$, the group $G_{\long}$ is $Sp(2)^n$ and $G_{\long} N(T) = Sp(2)^n \rtimes S_n$ [this group is table 8 fifth row $ p=1 $ case in https://arxiv.org/abs/math/0605784].
In type $F_4$, the group $G_{\long}$ is $Spin(8)$ and $ G_{\long} N(T)=Spin(8)\rtimes S_3 $ [see comments on https://math.stackexchange.com/questions/88330/the-weyl-group-of-f-4 for some details about this subgroup]
In $G_2$, the group $G_{\long}$ is $SU(3)$. $G_{\long} N(T)$ is $SU(3) \rtimes 2$, with $2$ acting by the unique outer automorphism of $ SU_3 $ (complex conjugation).
In general it seems that $$ G_{long}N(T)=G_{long}\rtimes Out(G_{long}) $$