Consider the following functional:
$$ E_k(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$
Theorem:
The Euler-Lagrange equation of $E_2$, is $A(\phi)=0$, where $A(\phi) \in \Gamma(\phi^*\TN)$ is defined by
$$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge
d\phi)\big)\bigg).$$
$$h_{\phi^*TN}:\phi^*TN \otimes \Lambda_2(\phi^*TN) \to \phi^*TN$$ is a linear map, which depend on the metric on $\N$, and is defined precisely below (see eq $(6)$, and replace $W$ with $\phi^*TN$).
Proof:
Let $\phi$ be a map $\M \to \N$, and $\phi_t:\M \to \N$ as smooth family, where $\phi_0=\phi$ and $\frac{\partial \phi_t}{\partial t}|_{t=0}:=V \in \Gamma(\phi^*(\TN))$. Then
$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\frac{1}{2}\int_{\M}\frac{\partial{}}{dt}|_{t=0} \| d\phi_t \wedge d\phi_t \|^2 \text{Vol}_{\M}= \int_{\M} \langle d\phi \wedge
d\phi, \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\rangle \text{Vol}_{\M}. \tag{1}$$
It is well-known that $\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0}=\nabla^{\phi^*(TN)}V \in \Gamma(T^*\M \otimes \phi^*(\TN))$.
Now,
$$\bigg(\nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\bigg)(X,Y)=$$
$$(\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(X) \wedge d\phi(Y)+d\phi(X) \wedge (\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(Y)=$$
$$ \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y)= $$
$$\big(\nabla V \wedge d\phi+d\phi \wedge \nabla V\big) (X,Y), \tag{2}$$
where $\nabla V \wedge d\phi+d\phi \wedge \nabla V \in \Omega^2\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$ is defined by the last equality.
Thus, we have obtained
$$ \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}= \nabla V \wedge d\phi+d\phi \wedge \nabla V. \tag{3}$$
Define also $\xi=V \wedge d\phi \in \Omega^1\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$.
Lemma: $d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)=\nabla V \wedge d\phi+d\phi \wedge \nabla V$.
Assuming the lemma, we combine equations $(1),(3)$ and get
$$ \frac{d}{dt}|_{t=0}E(\phi_t)=\int_{\M} \langle d\phi \wedge
d\phi, d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)\rangle \text{Vol}_{\M}=\int_{\M} \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge
d\phi),\xi\rangle \text{Vol}_{\M}. \tag{5}$$
To find the exact $E-L$ equations, one further step needs to be taken:
$$V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge
d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge
d\phi),V \wedge d\phi\rangle$$ is a linear functional in $V$, so it can be expressed as $V \to \langle V, A(\phi) \rangle_{\phi^*\TN}$,
where $A(\phi) \in \Gamma(\phi^*\TN)$. The E-L equation is $A(\phi)=0$.
We now turn to finding an explicit expression for this "representation":
The corresponding pointwise linear algebra situation is this:
We have two oriented $d$-dimensional inner product spaces $V,W$, together with maps $A \in \Hom(V,W),B \in \Hom(V,\Lambda_2(W))$, and we look for a bilinear map $$\psi: \Hom(V,W) \times \Hom(V,\Lambda_2(W)) \to W,$$ satisfying
$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}=\langle w, \psi(A,B) \rangle_W \, \text{ for every $w \in W$}$$
Proposition: With the notation as above, $\psi(A,B)=h_W\big(\tr_{V} (A \otimes B)\big)$ where $h_W:W \otimes \Lambda_2(W) \to W$ is defined by the linear extension of
$$ \tilde w \otimes (w_1 \wedge w_2) \to \langle \tilde w,w_2 \rangle w_1-\langle \tilde w,w_1 \rangle w_2. \tag{6}$$
Note $A \otimes B \in V^* \otimes V^* \otimes W \otimes \Lambda_2(W)$, so $\tr_{V} (A \otimes B) \in W \otimes \Lambda_2(W)$.
Proof:
It suffices to prove this for $A,B$ "pure" tensors, i.e $A=\alpha \otimes \tilde w,B=\beta \otimes (w_1 \wedge w_2)$, where $\alpha,\beta \in V^*,\tilde w,w_1,w_2 \in W$.
Now, on the one hand
$$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}= \langle \alpha \otimes (w \wedge \tilde w) ,\beta \otimes (w_1 \wedge w_2) \rangle_{\Hom(V,\Lambda_2(W))}=$$
$$ \langle \alpha , \beta \rangle_{V^*} \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}. $$
On the other hand
$$ \tr_{V} (A \otimes B)= \langle \alpha , \beta \rangle_{V^*} \tilde w \otimes (w_1 \wedge w_2).$$
Thus, it's enough to show
$$ \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}=\langle w , h_W\big(\tilde w \otimes (w_1 \wedge w_2)\big) \rangle_W,$$
but this nows follows directly form the definition of the induced inner product on $\Lambda_2(W)$, and the definition of $h_W$ (see $(6)$).
Using the above proposition, we deduce that $$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge
d\phi)\big)\bigg).$$
Best Answer
A. multi-dimensional state, one-dimensional time
Multi-dimensional extensions $x\in\mathbb{R}^n$ of the one-dimensional Hamilton-Jacobi-Bellman equations have been considered in Consistency of a Simple Multidimensional Scheme for Hamilton-Jacobi-Bellman Equations (2005).
B. multi-dimensional state, multi-dimensional time
For extensions where both state and time are multi-dimensional, $x\in\mathbb{R}^n$, $t\in\mathbb{R}_+^m$, see Multitime linear-quadratic regulator problem based on curvilinear integral (2009) (and several more recent papers on the multi-time Bellman principle by Constantin Udriste and co-workers).