I am interested in subgroups of $\text{Spin}(7)$ and identifying certain elementary properties that they have. Specifically relating to commutativity. Let me apologise at this point for my ignorance and sloppy presentation below. I know little about this subject, which has recently come up in my work.
I write down the extended Dynkin diagram $\widetilde B_3$ corresponding to the Lie algebra of $\text{Spin}(7)$
with appended long root $\alpha_0$, and short root $\alpha_3$.
Removing the short node $\alpha_3$ I get $A_3$ coresponding to the $SU(4)\cong \text{Spin}(6)$ subgroup. Cutting out $\alpha_1$ leaves something with a fold symmetry that gives the $G_2\leq \text{Spin}(7)$ subgroup inclusion.
On the other hand, if we take out $\alpha_2$ then we are left with three $A_1$'s. It's pretty clear that the pair given by $\alpha_0$, $\alpha_1$ are the two commuting $\text{Spin}(3)$-subgroups that together constitute the canonical $\text{Spin}(4)\leq \text{Spin}(7)$. The final $A_1$ corresponds to the short root $\alpha_3$, and is what is confusing me. My intuition is telling me that it is a third copy of $\text{Spin}(3)$, covering the $SO(3)\leq SO(7)$ which is complementary to the $SO(4)$ subgroup.
As such this third $A_1$ should be another $\text{Spin}(3)$ which commutes with the $\text{Spin}(4)$ identified previously. However, my understanding runs out here, and the presence of the half-length root indicates to me that what I have actually identified is in fact an $(S^3\times S^3\times S^3)/\mathbb{Z}_2$ subgroup containing a $\text{Spin}(4)$ subgroup which does not commute with the other factor.
Have I identified a commuting $\text{Spin}(3)$ and $\text{Spin}(4)$ subgroups, or is it rather the second option $(S^3\times S^3\times S^3)/\mathbb{Z}_2$?
Moreover, the diagram indicates to me the presence of three distinguished homotopy classes of maps $S^3\rightarrow \text{Spin}(7)$. The first two come from $\text{Spin}(4)$, with $\alpha_0$ corresponding to the canonical generator $i_0:S^3\cong \text{Spin}(3)\hookrightarrow \text{Spin}(7)$, and $\alpha_1$ its negative, $i_1=-i_0$. Since $\frac{|\alpha_0|^2}{|\alpha_3|^2}=2$, the corresponding homotopy class of $\alpha_3$ should be $i_3=2\cdot i_0$, which would follow from elementary properties of the Dynkin index. I'm not entirely sure this is what I want to see.
Am I correct about the identifications here?
I don't know if you still need an answer to this question, and I don't know enough about roots to answer the question using that approach, but I can confirm a positive answer to your questions. That is, there is (up to conjugacy) a unique subgroup of $Spin(7)$ which is locally isomorphic to $SU(2)^3$. This subgroup is isomorphic to $(SU(2))^3/\langle -(I,I,I)\rangle$. Second, the inclusions of the $SU(2)$ factors into $Spin(7)$ have Dynkin index $2$, $1$, and $1$.
I'll sketch proofs of all this below.
First, let's work in $SO(7)$. A finite-to-one homomorphism $f:SU(2)^3\rightarrow SO(7)$ is nothing but a real $7$-dim representation of $SU(2)^3$, so let's try to understand those.
We will need the following facts:
For each integer $n \geq 1$, there is a unique $n$-dimensional complex irreducible representation of $SU(2)$. It has real type if $n$ is odd, and symplectic type if $n$ is even.
Irreducible representations of a product of Lie groups are given by tensor products of irreducible representations of the factors. The tensor product of two real representations or two symplectic representations is real.
A representation $V$ is real iff it can be decomposed into irreps $V = \oplus V_i \oplus (W_j \oplus \overline{W}_j)$ where all $V_i$ are real and $\overline{W}_j$ indicates the complex conjugate of $W_j$.
Proposition: There is a unique real $7$-dimensional representation of $SU(2)^3$ with the property that the restriction to each $SU(2)$ factor is non-trivial.
Proof: I'll use notation $m\otimes n\otimes p$ to refer to the tensor product of the $m$, $n$, and $p$-dimensional irreps of $SU(2)$.
We first list all the irreps of $SU(2)^3$ which have dimension at most $7$.
They are, up to permutations, $1\otimes 1\otimes p$ for $p\leq 7$, $1\otimes 2\otimes 2$, $1\otimes 2\otimes 3$.
Existence: the representation $(3\otimes 1\otimes 1)\oplus (1\otimes 2\otimes 2)$ is real, and non-trivial on all three $SU(2)$ factors.
Uniqueness: We show no other combination works. If our $7$-dim representation has $1\otimes 2\otimes 3$ as a subrepresentation, then the complmentary subrepresentation has dimension $1$, so is trivial. Thus, such a representation is trivial on some $SU(2)$ factor, ruling out this case.
Likewise, if our $7$-dim representation contains $1\otimes 1\otimes p$ for $p\geq 4$, then the complementary subrepresentation has dimension $7-p < 4$, so can't be injective on the other two $SU(2)$ factors.
If our $7$-dim representation doesn't contain $1\otimes 2\otimes 2$, and is non-trivial on all $SU(2)$ factors, then, up to permutation, it must be $(2\otimes 1\otimes 1)\oplus(1\otimes 2\otimes 1)\oplus (1\otimes 1\otimes 2)\oplus (1\otimes 1 \otimes 1)$ or $(3\otimes 1\otimes 1)\oplus(1\otimes 2\otimes 1)\oplus (1\otimes 1\otimes 2)$. In both these cases, the representation is complex, not real.
So, our $7$-dim representation is either $(1\otimes 2\otimes 2)\oplus (2\otimes 1\otimes 1)\oplus (1\otimes 1\otimes 1)$, or it is $(1\otimes 2\otimes 2)\oplus (3\otimes 1\otimes 1)$. But the first of these is not real, and the second is the one I gave in the "existence" portion of the proof. $\square$
A result of Malcev (and I can try to get the precise reference if you need it) indicates equivalent representations have conjugate images in $SO(7)$. (The actual result is that this is true for all classical groups except $SO(even)$).
So, up to conjugacy, there is a unique homomorphism $f:SU(2)^3\rightarrow SO(7)$ with finite kernel. This is the one with image given by the block embedding of $SO(3)\times SO(4)$. (I'm sure you were already aware that there were commuting $SO(3)$ and $SO(4)$ in $SO(7)$. The point of all this is that we've established that any subgroup of $SO(7)$ which is locally $SU(2)^3$ is this one, up to conjugacy. Thus, any subgroup of $Spin(7)$ which is locally $SU(2)^3$ must be a lift of this one.)
Now we'll try to lift all this to $Spin(7)$. Of course, the map $f$ lifts since $SU(2)^3$ is simply connected, but what is the kernel of the lift $\tilde{f}$? Obviously, $\ker \tilde{f}\subseteq \ker f = \langle (-I,I,I), (I,-I,-I)\rangle$, but which subgroup is it?
I claim that it's $\langle -(I,I,I)\rangle$, as you guessed. To see this, simply observe that the maps $SO(3),SO(4)\rightarrow SO(7)$ can't lift because they're non-trivial on $\pi_1$. Thus, the $\ker \tilde{f}$ is either trivial or it's $\langle -(I,I,I)\rangle$, but it can't be trivial because $\pi:Spin(7)\rightarrow SO(7)$ has a kernel of order $2$, while $f$ has a kernel of order $4$.
What about the corresponding elements of $\pi_3(SO(7))$ (which is, of course, isomorphic to $\pi_3(Spin(7))$?
Well, for the map $SU(2)\rightarrow SO(3)\times \{I\}\subseteq SO(3)\times SO(4)\subseteq SO(7)$, the inclusion $SO(3)\subseteq SO(7)$ has Dynkin index $2$.
To see this, first observe that $SO(5)/SO(3)$ is the unit tangent bundle of $S^4$, $T^1S^4$, so $H^4(SO(5)/SO(3))\cong \mathbb{Z}/2\mathbb{Z}$. On the other hand, since $Spin(5) = Sp(2)$, we can rewrite this as $SO(5)/SO(3) = Sp(2)/\Delta Sp(1)$. Since the symplectic groups are $2$-connected, we see that $\pi_3(Sp(2))/\pi_3(Sp(1))\cong \pi_3(T^1 S^4)\cong H_3(T^1 S^4)\cong H^4(T^1 S^4)$, so the Dynkin index of $SO(3)$ in $SO(5)$ is $2$. On the other hand, the Dynkin index of $SO(5)$ in $SO(7)$ is $1$ since $SO(7)/SO(5) = Spin(7)/Spin(5) = T^1 S^6$ which is $4$-connected.
Lastly, the Dynkin index of either $SU(2)\subseteq SO(4)$ in $SO(7)$ is $1$. To see, note that using the previous paragraph, it is sufficient to show that the Dynkin index of either $SU(2)$ in $SO(5)$ is $1$. Regardless of which of the two $SU(2)$ we pick, $SO(5)/SU(2)$ is double covered by $Sp(2)/SU(2) = S^7$, which is $6$-connected. So the Dynkin index of $SU(2)$ in $SO(5)$ is $1$.