The conclusion is that the ADN weighted principal symbol $\sigma_{\text{pr},w}(\xi) = D_{[d]}$ is just the leading homogeneous term with respect to a grading given by the weights $a_i = s_i$, $b_j = -t_j$ and $d=0$. But as we have seen earlier, the $d=0$ condition is easily relaxed just by shifting the weights. What is interesting about the definition of the weighted principal symbol as the leading homogeneous term is that, for a fixed choice of weights, it is easy to see that it is well-defined and transforms covariantly in $\xi$ under changes of the $(x^l)$ coordinates (subleading terms in the transformation of highest order derivatives in each component of $D_{ij}$ only contribute to subleading terms $D_{[c]}$, while the highest derivatives transform covariantly).

The usual definition of the principal symbol can be recovered by setting all the weights to zero, $a_i = b_j = 0$. However, the simple example $$ \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} = \begin{pmatrix} \Delta^3 & 0 & 0 \\ 0 & \Delta^2 & 0 \\ 0 & 0 & \Delta \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} $$ shows that such a default choice picks up only the highest order $\Delta^3$ term, making the principal symbol degenerate and hence, strictly speaking, non-elliptic. However, we can plainly see that the system is just a combination of three independent elliptic equations of different orders. Selecting the weights/gradings to be $b_j = 0$ and $a_1 = 6$, $a_2 = 4$ and $a_3 = 2$ allows the weighted principal symbol to pick up all the components (in this example there are no subleading terms) and become non-degenerate. The ADN definition of ellipticity simply generalizes this idea. The operator $D$ is ADN elliptic if there exists a choice of weights/gradings such that its weighted principal symbol (leading homogeneous term) is invertible for $\xi\ne 0$.

Finally, to the question whether there exists some automatic way of choosing the weights, in general I don't know. Certainly, it's possible that an operator is not elliptic with respect to any weights. Perhaps there is a way to decide when an operator can be made elliptic with respect to some choice of weights, but I don't know it of the top of my head. Possibly this question can be answered within commutative algebra and the theory of graded modules.

The conclusion is that the ADN weighted principal symbol $\sigma_{\text{pr},w}(\xi) = D_{[d]}(\xi)$ is just the leading homogeneous term with respect to a grading given by the weights $a_i = s_i$, $b_j = -t_j$ and $d=0$. But as we have seen earlier, the $d=0$ condition is easily relaxed just by shifting the weights. What is interesting about the definition of the weighted principal symbol as the leading homogeneous term is that, for a fixed choice of weights, it is easy to see that it is well-defined and transforms covariantly in $\xi$ under changes of the $(x^l)$ coordinates (subleading terms in the transformation of highest order derivatives in each component of $D_{ij}$ only contribute to subleading terms $D_{[c]}(\xi)$, while the highest derivatives transform covariantly).

The usual definition of the principal symbol can be recovered by setting all the weights to zero, $a_i = b_j = 0$. However, the simple example $$ \tag{$*$} \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} = \begin{pmatrix} \Delta^3 & 0 & 0 \\ 0 & \Delta^2 & 0 \\ 0 & 0 & \Delta \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} $$ shows that such a default choice picks up only the highest order $\Delta^3$ term, making the principal symbol degenerate and hence, strictly speaking, non-elliptic. However, we can plainly see that the system is just a combination of three independent elliptic equations of different orders. Selecting the weights/gradings to be $b_j = 0$ and $a_1 = 6$, $a_2 = 4$ and $a_3 = 2$ allows the weighted principal symbol to pick up all the components (in this example there are no subleading terms) and become non-degenerate. The ADN definition of ellipticity simply generalizes this idea. The operator $D$ is ADN elliptic if there exists a choice of weights/gradings such that its weighted principal symbol (leading homogeneous term) is invertible for $\xi\ne 0$.

Finally, to the question whether there exists some automatic way of choosing the weights, in general I don't know (but see UPDATE below). Certainly, it's possible that an operator is not elliptic with respect to any weights (see (a) below). Perhaps there is a way to decide when an operator can be made elliptic with respect to some choice of weights, but I don't know it of the top of my head (see (b) below). Possibly this question can be answered in general within commutative algebra and the theory of graded/filtered modules.

UPDATE: For completeness, I'll summarize here the main points of the nice paper of Volevich cited in the answer of Michael Renardy, which addresses (a) when the weights might exist, (b) how to find them, and (c) how unique are they. Throughout, let $T(\partial)$ be a square matrix of linear differential operators and $T(\xi)$ the corresponding polynomial matrix, where we have substituted $\partial_l = \partial/\partial x^l \mapsto \xi_l$ after, say, moving all the derivatives to the right.

(a) Recall the determinant formula $\det T(\xi) = \sum_\pi (-)^{|\pi|}\prod_i T_{i\pi(i)}(\xi)$ with summation over permutations $\pi$. Let $r(T) = \max_\pi \sum_i d_{i\pi(i)}$ denote the largest polynomial degree that appears in the determinant formula; let's call it the formal determinant degree. When there are no cancellations among the highest order terms, the formal degree $r(T)$ coincides with the actual polynomial degree of $\det T(\xi)$. If such cancellations do occur, the actual determinant degree may be lower. These are the cases that we exclude from consideration, since then there might not exist a set of weights that have all the nice properties that we might want of them. What it means is that in such cases the bases for the equations and the dependent variables, with respect to which the matrix elements $T_{ij}(\partial)$ are expressed are not well-adapted to the PDE system. To get around the problem, one would have find a better adapted form of the PDE (mixing equations, deriving integrability conditions, etc.). When people talk about putting equations into an involutive form that's what they are after.

(b) From now on, we can assume that $r(T)$ coincides with the polynomial degree of $\det T(\xi)$. The goal now is to choose some weights $a_i$, $b_j$ such that the weighted principal symbol of $T(\xi)$ reproduces $\det \sigma_{\text{pr},w}(\xi) = \sigma(\det T(\xi))$, where the right-hand side is the "principal symbol" of $\det T(\xi)$, which is the degree-$r(T)$ homogeneous piece of $\det T(\xi)$. Incidentally, by defining the "correct weights" in this way, one can already check the ADN condition for ellipticity without knowing what the weights are explicitly (which echos Deane Yang's comment). Namely, $$ \forall \xi\ne 0 : \exists \sigma_{\text{pr},w}(\xi)^{-1} ~ \text{(ADN ellipticity)} \iff \forall \xi\ne 0 : \sigma(\det T(\xi)) \ne 0 . $$

Now, to actually find the weights, we need the magic step from Volevich, which reformulates $r(T)$ as the optimal value of a linear program: $$\begin{aligned} r(T) &= \max_\pi \sum_i d_{i\pi(i)} \\ &= \max_\pi \sum_{ij} d_{ij} \pi_{ij} \\ &= \max_{p \in \operatorname{conv} \{\pi\}} \sum_{ij} d_{ij} p_{ij} \\ &= \max \left\{ \sum_{ij} d_{ij} p_{ij} \mid p_{ij} \ge 0, \sum_i p_{ij} = 1, \sum_j p_{ij} = 1 \right\} , \end{aligned}$$ where we have at first interpreted a permutation $\pi$ as a corresponding permutation matrix $\pi_{ij}$, then using the linearity of the function being maximized we have replaced $\pi_{ij}$ by an arbitrary matrix $p_{ij}$ in the convex hull of the permutation matrices, which happen to consist of the so-called doubly stochastic matrices (Birkhoff-von Neumann theorem). From this observation, the desired weights can be obtained by appealing to a general fact from linear programming: $r(T)$ is also the optimal value of a dual linear program. Unfolding the definition of the dual linear program gives $$ r(T) = \min \left\{ \sum_i a_i - b_i \mid a_i - b_j \ge d_{ij} \right\} , $$ where we now optimize over the dual variables $a_i$, $b_j$ (or $a_i$, $(-b_j)$ in more standard notation), which we immediately realize as the desired weights! The weights exist because this dual linear program achieves its optimal value, which is true because the original linear program over $p_{ij}$ achieves its optimal value, which in turn is due to the compactness of the set of doubly stochastic matrices (as they are the convex hull of the finitely many permutation matrices).

(c) As was already noted earlier, a given set of weights produces the same weighted principal symbol under the constant shift $(a_i, b_j) \mapsto (a_i+c, b_j+c)$. In the extreme example of a diagonal system $(*)$, the pair of weights for each diagonal entry can be similarly shifted independently from each other. One more thing that Volevich proves is that either the weights are unique (up to the above global shift) or the weighted principal symbol can be row/column permuted into diagonal blocks and the weights can be shifted (roughly) independently within each block. (I haven't fully absorbed that part of the proof, but I think it means that the shift within each diagonal block can be made independently within some range, determined by the $d_{ij}$ degrees from the off-diagonal blocks that don't contribute to the weighted principal symbol directly.)

There is a notion of a homogeneous map between graded vector spaces. Namely $T \colon V \to U$ is homogeneous of degree $c$ if $T(V_b) \subseteq U_{a=b+c}$. Written as a matrix $T = (T_{ij})$, only those components $T_{ij}$ can be non-zero for which $a_i - b_j = c$. Of course, any linear map can be written as a sum of homogeneous pieces, $T = \sum_c T_{(c)}$. When the components of $T$ are not just numbers but polynomials in $\xi_l$, conventionally the variables $\xi_l$ are all given weight 1 and the notion of homogeneity changes slightly. Namely, $T$ is (polynomially) homogeneous of degree $c$ if its components $T_{ij}$ are homogeneous polynomials, say of degrees $d_{ij}$, and are non-vanishing only when $a_i - b_j - d_{ij} = c$. An arbitrary linear operator with polynomial entries with can again be expanded in homogeneous pieces, $T = \sum_{c} T_{[c]}(\xi)$, where each polynomially homogeneous pieces can again be expanded as $T_{[c]}(\xi) = \sum_{|\alpha|=c} T_{(c-|\alpha|)} \xi^\alpha$ with respect to the first notion of homogeneity. What I ad-hoc called polynomial homogeneity coincides with the notion of homogeneity when we consider $U$ and $V$ are graded modules over the polynomial ring $\mathbb{R}[\xi]$, where the $\xi_l$ variables all have weight 1.

There is a notion of a homogeneous map between graded vector spaces. Namely $T \colon V \to U$ is homogeneous of degree $c$ if $T(V_b) \subseteq U_{a=b+c}$. Written as a matrix $T = (T_{ij})$, only those components $T_{ij}$ can be non-zero for which $a_i - b_j = c$. Of course, any linear map can be written as a sum of homogeneous pieces, $T = \sum_c T_{(c)}$. When the components of $T$ are not just numbers but polynomials in $\xi_l$, conventionally the variables $\xi_l$ are all given weight 1 and the notion of homogeneity changes slightly. Namely, $T$ is (polynomially) homogeneous of degree $c$ if its components $T_{ij}$ are homogeneous polynomials, say of degrees $d_{ij}$, and are non-vanishing only when $a_i - b_j - d_{ij} = c$. An arbitrary linear operator with polynomial entries which can again be expanded in homogeneous pieces, $T = \sum_{c} T_{[c]}(\xi)$, where each polynomially homogeneous pieces can again be expanded as $T_{[c]}(\xi) = \sum_{|\alpha|=c} T_{(c-|\alpha|)} \xi^\alpha$ with respect to the first notion of homogeneity. What I ad-hoc called polynomial homogeneity coincides with the notion of homogeneity when we consider $U$ and $V$ are graded modules over the polynomial ring $\mathbb{R}[\xi]$, where the $\xi_l$ variables all have weight 1.