For anyone who cares, I have figured out the answer to the problem. If one labels the map $\alpha\circ p_1=\beta\circ p_2:A\times_X B\to X$ as $h$, then we have two commutative triangles making up the front face of the cube. Take the triangle $A\times_X B\to A\to X$. One forms pullbacks $h'$ and $\alpha'$ of $h$ and $\alpha$ respectively along $f$, and then two-pullbacks gives a map $q_1:P\to A'$ such that $\alpha'\circ q_1=h'$ and such that left side of the cube is a pullback. Do the same for the other triangle on the front face. Then one evidently gets commutativity of the entire cube, and every face save for the back face (the one we care about) is a pullback square. Now, by two-pullbacks, if we show that the rectangle formed by the back and bottom faces is a pullback, then since the bottom face is a pullback, the back face must be as well. But since the diagram commutes, this rectangle is the same as the rectangle formed by the top and front faces. Since these faces are each pullback squares, the rectangle is as well, and two-pullbacks gives that the back face is indeed a pullback.
Working through this problem has also made it evident to me why the commutative triangle lemma should be interpreted the way it is. In this problem, we assume that the category $\mathbf{C}$ has all pullbacks, but in this problem we are only pulling back along $f$. In order to get anywhere in this problem, i.e. show that a face which doesn't involve $f$ is a pullback, we need some way of concluding that some face which doesn't involve $f$ is a pullback. The commutative triangle lemma states that if one has a commutative triangle and pulls back the only two maps that one can pull back along $f$, then there is a map completing the triangle, that is, which makes a very much desired "commutative triangular prism" in which the faces are pullbacks.
If there are any flaws in my understanding someone should let me know, but as of now I feel that all of the gaps in my understanding have been filled.
I think you've seen why it ought to be the intersection, but for the sake of completeness I'll be especially painstaking about it.
Let's just use $i:E\rightarrowtail X$ for the subobject classified by $F\circ\langle\phi,\psi\rangle$, and $i_A,i_B$ be the inclusion morphisms of $A,B$, respectively, into $X$.
By the lefthand commutative square in the above diagram, we have that $\phi\circ i=\top\circ !_E$ (so $i$ factors through $i_A$) and $\psi\circ i=\top\circ !_E$ (so $i$ factors through $i_B$). It follows from this (along with the fact that $i$ is a monomorphism) that $E$ is a subobject of both $A$ and $B$, and this is our first clue that we're looking at what ought to be an intersection. Let's denote by $j_A,j_B$ the respective inclusions of $E$ into $A,B$; note, in particular, that $i_Aj_A=i=i_Bj_B$.
Now let $f:Q\to A$ and $g:Q\to B$ be morphisms such that $i_Af=i_Bg$; we wish to show that there is a unique map $h:Q\to E$ with $j_A\circ h=f$ and $j_B\circ h=g$; that is, we want to show that $E$ is the pullback of $i_A$ and $i_B$.
Denote by $\alpha$ the morphism $i_Af=i_Bg:Q\to X$. Then $$\phi\circ\alpha=\phi\circ i_A\circ f=\top\circ !_A\circ f=\top\circ!_Q$$ and $$\psi\circ\alpha=\psi\circ i_B\circ g=\top\circ !_B\circ g=\top\circ!_Q,$$ meaning that $\langle\phi,\psi\rangle\circ\alpha=\langle\top,\top\rangle\circ !_Q$. Because that left hand square is a pullback, there is a unique morphism $h:Q\to E$ with $i\circ h=\alpha$. This equality gives us both $$i_A\circ j_A\circ h=i\circ h=\alpha=i_A\circ f$$ from which we have $j_Ah=f$ because $i_A$ is a monomorphism, and $$i_B\circ j_B\circ h=i\circ h=\alpha=i_B\circ g$$ from which we have $j_B\circ h=g$.
Thus we've shown that $E$ satisfies the universal property of the intersection of $A$ and $B$, using only that the squares in the pictured diagram are pullbacks.
Best Answer
This is just a special case of the more general property of limits to commute with each other, perhaps that is why you had a hard time finding a reference for this.