Generalization of categorical product

The $F(\pi_i) = id$ doesn't really fit the form of a universal property. If we drop that constraint, we can present your universal property in terms of representability $$\mathcal C(Y, X_1 \times_F X_2) \cong \{(f_1,f_2)\in\mathcal C(Y, X_1)\times\mathcal C(Y, X_2)\mid F(f_1) = F(f_2) \}$$ natural in $Y$. Note that this still guarantees that $F(\pi_1)=F(\pi_2)$. $\pi_1$ and $\pi_2$ are the components we get by choosing $Y=X_1 \times_F X_2$ and considering the image of $id_{X_1 \times_F X_2}$.

This doesn't really strike me as anything natural. Nevertheless, we can massage it a bit to get something maybe a little bit nicer if still not terribly natural. We can note that the representation can seen as a special case for the hom-set of a comma category. In general, letting $H : \mathcal C \to \mathcal E$ and $K : \mathcal D \to \mathcal E$, the comma category, $H \downarrow K$, has as object triples $(C, \alpha, D)$ where $C$ is an object of $\mathcal C$, $D$ is an object of $\mathcal D$, and $\alpha : H(C) \to K(D)$. For morphisms, we have

$$(H \downarrow K)((C, \alpha, D), (C', \alpha', D')) = \{(f_1, f_2)\in\mathcal C(C, C')\times\mathcal D(D, D')\mid \alpha' \circ H(f_1) = K(f_2) \circ \alpha \}$$

We immediately see that choosing $H = K = F$ and $\alpha = id_{F(Y)}$ and $\alpha' = id_{F(X_1)} = id_{F(X_2)}$ we get the representation above. More precisely, we want to consider $$(F \downarrow F)((Y, id_{F(Y)}, Y), (X_1, id_{F(X_1)}, X_2))$$ This should be functorial in $Y$, so we define the functor $I_F(Y) = (Y, id_{F(Y)}, Y)$ and $I_F(g) = (g,g)$ which satisfies the constraint to be a morphism which just reduces to $F(g) = F(g)$ in this case. We can now write: $$\mathcal C(Y, X_1 \times_F X_2) \cong (F \downarrow F)(I_F(Y), (X_1, id_{F(X_1)}, X_2))$$ This would almost be in the form of an adjunction if only we considered an arbitrary morphism (in $F \downarrow F$) as the codomain. So let's do that.

$$\mathcal C(Y, X_1 \times_\alpha X_2) \cong (F \downarrow F)(I_F(Y), (X_1, \alpha, X_2))$$

natural in $Y$ and $(X_1, \alpha, X_2)$, i.e. $I_F \dashv ({-})\times_{({-})}({-})$. Of course, for this to be meaningful we'd need to define the functor $(X_1, \alpha, X_2) \mapsto X_1 \times_\alpha X_2$. Nevertheless, this transforms the question into looking for a right adjoint to $I_F$. The universal property would be the same as the one you list (minus the $F(\pi_i) = id$ constraint) except you'd have $\alpha \circ F(f_1) = F(f_2)$ which breaks $F(f_1)=F(f_2)$ and seems slightly more natural to me. Of course, you can choose $\alpha=id_{F(X_1)}=id_{F(X_2)}$ whenever $F(X_1) = F(X_2)$.