Suppose $\mathfrak{g}$ is a complex reductive Lie algebra and $\theta$ is an automorphism of order $2$. Let $\mathfrak{g} = \mathfrak{g_0} \oplus \mathfrak{g}_1$ be the corresponding $\mathbb Z_2$-gradation. Cartan defined the concept of Cartan subspace and the Weyl group of the symmetric pair $(\mathfrak{g}, \mathfrak{g_0})$. By definition a Cartan subspace of the pair $(\mathfrak{g}, \mathfrak{g_0})$ is a subspace $\mathfrak{c}$ of $\mathfrak{g}_1$ which is a maximal commutative subspace consisting of semisimple elements and the Weyl group is by definition is the quotient $N_{\mathfrak{g}_0}(\mathfrak{c})/Z_{\mathfrak{g}_0}(\mathfrak{c})$. What is a Cartan subspace and the Weyl group for the pair $(\mathfrak{sl}_{2n}, \mathfrak{sp}_{2n})$ ?. Is there a canonical choice for $\mathfrak{c}$ in this case ? What I understood from different sources is that the Weyl group is a subgroup/quotient of $S_{2n}$. Can it be described explicetely ?

For a general symmetric space (let’s say over $ \mathbb{C}$ for simplicity) to obtain a Cartan subspace, take a Cartan subalgebra $\mathfrak{t}$ of $\mathfrak{g}$ that is stable under the involution $\theta$. This decomposes $$ \mathfrak{t}=\mathfrak{t}_0\oplus \mathfrak{t}_1 $$ in terms of the eigenspaces of $\theta$. Choose such a Cartan so that the fixed subspace $\mathfrak{t}_0$ is of minimal dimension. The $-1$ eigenspace $\mathfrak{t}_1$ gives a Cartan subspace of the pair $(\mathfrak{g},\mathfrak{g}_0)$. Over $\mathbb{C}$, tori with this property (referred to as maximally $\theta$-spit in the literature) are characterized as those lying in a Borel subalgebra $\mathfrak{b}$ such that $\theta(\mathfrak{b})$ is the opposite Borel containing $\mathfrak{t}$: that is, $$ \mathfrak{b}\cap\theta(\mathfrak{b})=\mathfrak{t}. $$

In the case you mention, take the diagonal Cartan of $\mathfrak{sl}(2n)$ and the involution $$\theta(X)=-JX^TJ,$$ where $J$ is the standard "opposite diagonal" symplectic form.

As for the Weyl group, a result of Luna shows that if $W$ is the Weyl group of $(\mathfrak{g},\mathfrak{t})$ with $\mathfrak{t}=\mathfrak{t}_0\oplus \mathfrak{t}_1$ as above, consider the two subgroups $$ W_1=\{w\in W: w(\mathfrak{t}_1)=\mathfrak{t}_1\} $$ and $$ W_2=\{w\in W_1: w|_{\mathfrak{t}_1}\equiv 1\}, $$ then $W(\mathfrak{g}_0,\mathfrak{t}_1)\cong W_1/W_2$.