Comparison of $I$ with correlation coefficients is good, but it has its limits. This answer uncovers what those limits are. It derives a tight upper bound for $|I|$ in terms of the weights matrix $W$ and shows that in ordinary applications, where $W$ is symmetric and row-normalized, this bound is $1$ (or less). For the extraordinary weights matrix in the question (which has some zero rows, making it impossible to row-normalize), it shows that $X=\pm(3,-1,-1,-1)$ achieves the most extreme value of $|I|$ possible, which is $3$.
Analysis: Simplifying the Formula for $I$
The repeated appearances of $x_i - \bar x$ in the formula, along with the sum of their squares in the denominator, invite us to re-express $I$ in terms of the standardized variates
$$z_i = \frac{(x_i - \bar x)}{\sqrt{\sum_{i=1}^n (x_i - \bar x)^2}};\quad \bar x = \frac{1}{n}\sum_{i=1}^n x_i,$$
which can be assembled into the column vector $z = (z_1, \ldots, z_n)^\prime$. Because the variables have been standardized, $z$ is a unit vector
$$|z|^2 = z_1^2 + \cdots + z_n^2 = 1.$$
(This differs slightly from the usual statistical concept of "standardization," which recenters and rescales $x$ to have length $\sqrt{n}$.)
Further simplicity is achieved by renormalizing the weights $\omega_{ij}$ so they sum to unity:
$$\sum_{i,j} \omega_{ij} = 1.$$
When $S_0 = \sum_{i,j}\omega_{ij} \ne 0$, this can always be done by dividing every weight by $S_0$. (When the sum of the weights is zero, the formula for $I$ is undefined in the first place.) The renormalized weights form a matrix $W = (\omega_{ij} / S_0) = (w_{ij})$.
In terms of $z$ and $W$,
$$\eqalign{
I &= \sum_{i,j} \frac{n \omega_{ij}}{\sum_{i,j} \omega_{ij}} \left(\frac{(x_i-\bar x)}{\sqrt{\sum_i(x_i-\bar x)^2}}\right)\left(\frac{(x_j-\bar x)}{\sqrt{\sum_i(x_i-\bar x)^2}}\right) \\
&= \sum_{i,j} n w_{ij}z_i z_j = z^\prime (n W z).
}$$
This is identical to the formula for the Pearson correlation
$$\rho(z, w) =z^\prime w$$
between any two standardized variables $z$ and $w$.
Comparison to Correlation Coefficients
The difference between $I$ and a correlation is that $w = nWz$ might not be standardized: $|nWz| = \sqrt{(nWz)^\prime (nWz)}$ is not necessarily equal to $1$. To standardize, we must first subtract the mean of $w = nWz$,
$$\bar{w} = \frac{1}{n}\mathbf{1}^\prime n W z = \mathbf{1}^\prime W z,$$
from each component of $n W z$. Let $V$ be the matrix for this transformation,
$$V:z \to Vz = nWz - \mathbf{1}\bar w = \left(n\mathbb{I}_n - \mathbf{1}\mathbf{1}^\prime\right) W z.$$
(The vector $\mathbf{1}$ is the column vector $(1,1,\ldots,1)^\prime$, so the matrix $\mathbf{1}\mathbf{1}^\prime$ is the $n\times n$ matrix all of whose entries are ones.)
By standardizing $Wz$ in this manner--subtracting $\bar w$ and dividing by the length of the resulting centered vector--we get an honest correlation coefficient whose size cannot exceed $1$ (by virtue of the Cauchy-Schwarz Inequality):
$$1 \ge |\rho(z, Wz)| = \frac{|z^\prime Vz|}{|Vz|}.$$
This can be related to $I$ because $\mathbf{1}^\prime z = 0$, whence $z^\prime V z = z^\prime W z$. Clearing the denominator in the preceding inequality gives
$$|I| = |z^\prime (nW z)| = n|z^\prime V z| \le |V z|.$$
Equality will hold if and only if $Vz \ne 0$ and $Vz$ is parallel to $z$: that is, when $z$ is an eigenvector of $V$ with nonzero eigenvalue. When such an eigenvector exists (and it almost always will), this shows that
The size of $I$ is bounded by the largest eigenvalue (in absolute value) of $V = (n\mathbb{I} - \mathbf{1}\mathbf{1}^\prime)W$ and it can attain this bound.
Whenever we can find a basis in which $V$ is diagonal the largest absolute entry of $V$ is its largest eigenvalue.
In the question the unit-sum-normalized version of the weights matrix is
$$W = \pmatrix{1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0}$$
whence
$$V = \pmatrix{ 3&0&0&0 \\ -1&0&0&0 \\ -1&0&0&0 \\ -1&0&0&0 }.$$
It is simple to compute that the largest value attained by $|Vz|$ among all centered unit vectors $z$ (that is, $\mathbf{1}^\prime z = 0$ and $|z|=1$) is $3$, uniquely attained at $z=\pm(3,-1,-1,-1)/\sqrt{12}$. Consequently, all we can hope in general for this particular weights matrix $W$ is that $|I| \le 3$.
Resolution of the Original Question
Ordinarily, $W$ has nonnegative entries that originally are row standardized to sum to unity. This automatically makes the sum of each row in the unit-sum-normalized version of $W$ equal to $1/n$. Consequently $n W \mathbf{1} = \mathbf{1}$, making $\mathbf{1}$ an eigenvector of $n W$ with eigenvalue $1$.
It is also well-known that in this case if $W$ is symmetric, all other eigenvectors $z$ will be orthogonal to $\mathbf{1}$--that is, $z$ will be zero-centered--and will have eigenvalue $\lambda$ between $0$ and $1$. Considering any such unit $z$,
$$z^\prime V z = z^\prime\left(n\mathbb{I}_n - \mathbf{1}\mathbf{1}^\prime\right)W z = z^\prime (nW z) = z^\prime (\lambda z) = \lambda z^\prime z = \lambda \le 1.$$
In this ordinary application of Moran's I we may conclude that $\max{|I|} = 1$, as expected. But when $W$ is not symmetric and row-standardized--or cannot possibly be row-standardized because one or more rows are entirely zero--then the maximum has to be computed using the general result attained previously.
Best Answer
Moran's I statistic is used to explore a specific type of spatial clustering: whether high values are located in proximity to other high values and whether low values are located in proximity to other low values.
The trick then is 1st to get a sense of what you mean by proximity and 2nd formulating this mathematically. This idea of proximity will depend on the what type of observations (attributes) you are working with and what type of questions you have in mind.
For example, for human beings proximity could mean the distance needed to have a chat. So, if you wanted to know whether high income people like to chat with other high income people at your cocktail party, you could formulate proximity by using binary weights where 1 is defined by 2 people being within 3 feet of each other. To see whether house prices are spatially correlated you could define proximity as when 2 houses are neighbors or perhaps if two houses are on the same block or if 2 houses are within sight of one another etc etc.
Basically, you need a hypothesis of proximity that is based on some of your prior common sense ideas or expert knowledge of why 2 objects that are close to one another are more associated than 2 objects that are far from one another.
Moran's I can then be seen as a test of your hypothesis of how your notion of proximity structures high values next to one another on the landscape.