I'm not aware of any widely understood set of formal criteria that a purported measure of dispersion has to meet<sup>†</sup>—perhaps it's more that *dispersion* is a vague, pre-quantitative, notion that has motivated & can apply to various precisely defined measures. The assertion that "a measure of dispersion can, in the true sense, be regarded as the proper measure of dispersion if the measure is based on the deviations between all pairs of data" is intriguing; but its author doesn't provide any rationale (or any reference).<sup>‡</sup> This criterion would not in fact rule out the standard deviation: for a sample ${x_1, \ldots, x_n}$, the sum of squared deviations from the mean is equal to the mean of squared pairwise differences,
$$\sum_{i=1}^n{\left(x_i- \frac{\sum_{i =1}^n x_i}{n}\right)^2} = \frac{\sum_{i=1}^n\sum_{j = i + 1}^n(x_i - x_j)^2}{n}$$
(see https://stats.stackexchange.com/a/225758/17230).
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† @NickCox made me aware in short order that [Bickel & Lehmann (1976), *Ann. Statist.*, **4**, 6, "Descriptive statistics for nonparametric models. III. Dispersion"](https://projecteuclid.org/euclid.aos/1176343648), propose such a set of criteria. A measure of dispersion must (1) be location-invariant, (2) be scale-equivariant, & (3), given distributions $F_X$ symmetric about $\mu_X$ & $F_Y$ symmetric about $\mu_Y$ where $|X-\mu_X|$ stochastically dominates $|Y-\mu_Y|$, be greater (or at least equal) for $F_X$ than for $F_Y$.
That's quite intuitive, but to propound criteria for a measure of dispersion for an asymmetric distribution, you'd have to privilege one particular measure of central tendency—conjuring the spectre of *mean-dispersion*, *median-dispersion*, &c. [Bickel & Lehmann, (1976), in *Contributions to Statistics: Jaroslav Hajek Memorial Volume*, "Descriptive statistics for nonparametric models, IV. Spread"](https://link.springer.com/content/pdf/10.1007/978-1-4614-1412-4_45.pdf) take a different tack: noting that for symmetric distributions $X$ is more dispersed than $Y$ when
$$
\begin{align}
F_X^{-1}(v) - F_X^{-1}\left(\frac{1}{2}\right) &\leq F_Y^{-1}(v) - F_Y^{-1}\left(\frac{1}{2}\right)\quad \forall\, v \leq \frac{1}{2}\\
F_X^{-1}(v) - F_X^{-1}\left(\frac{1}{2}\right) &\geq F_Y^{-1}(v) - F_Y^{-1}\left(\frac{1}{2}\right)\quad \forall\, v \geq \frac{1}{2}
\end{align}
$$
, they decide that for asymmetric distributions $X$ is more *spread out* than *Y* when
$$
F_X^{-1}(v) - F_X^{-1}(u) \geq F_Y^{-1}(v) - F_Y^{-1}(u)\quad \forall\, u < v
$$
So any two quantiles of $X$ are farther apart than the corresponding quantiles of $Y$; the same probability mass is spread more thinly.
See the papers for proofs that standard deviation is both a measure of dispersion (for symmetric distributions) & of spread (for both symmetric & asymmetric distributions).
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‡ I see the rationale now: it's equivalent to saying that a dispersion measure must be invariant to translation &/or reflection.