Does this article use an inappropriate chi-square test or some other test of significance? I am trying to understand a result from an academic article analysing paragraphs of text. It reports a significance of p < .001, which is obviously strongly significant. Unfortunately, I cannot replicate this calculation. I assume that this is due to my limited knowledge of statistics, but I would like to rule out author error.
In the text, the author reports that any of the features of interest was more likely to occur than no feature of interest: 48 paragraphs with features, 12 features without, p < .001. No further details are given. This is the finding that I am interested in. 
There is a table that reports the occurrence of all the features across multiple genres of text. I cannot replicate the calculation for degrees of freedom displayed here and a chi-square test seems inappropriate for such low values. (The variables have been anonymized but the values are unchanged.) 

What significance test might have provided the p < .001 finding?
How could the chi-square value for the table have been calculated?
Is there an obvious justification for using a chi-square test with such low values?
 A: The report of 8 df implies to me 5 rows by 3 columns in the table used, and a few moments' work reproduces the result, with the easy guess that "Some feature" is just a secondary tabulation not echoed in the chi-square (which would indeed be absurd double counting). I use Stata here, but the calculations should be trivial in your favourite software. If not, change your favourite software. 
. tabchii 9 2 8 \ 2 3 2 \ 1 6 2 \ 5 3 5 \ 3 6 3, replace

          observed frequency
          expected frequency

-------------------------------
          |         col        
      row |     1      2      3
----------+--------------------
        1 |     9      2      8
          | 6.333  6.333  6.333
          | 
        2 |     2      3      2
          | 2.333  2.333  2.333
          | 
        3 |     1      6      2
          | 3.000  3.000  3.000
          | 
        4 |     5      3      5
          | 4.333  4.333  4.333
          | 
        5 |     3      6      3
          | 4.000  4.000  4.000
-------------------------------

12 cells with expected frequency < 5

         Pearson chi2(8) =  11.5941   Pr = 0.170
likelihood-ratio chi2(8) =  12.2947   Pr = 0.139

Those worried by the large number of observed frequencies below 5 should be cheered by none being below 1 and by strong consistency with a Fisher's exact test. 
. tab row col [w=observed], exact
(frequency weights assumed)

Enumerating sample-space combinations:
stage 5:  enumerations = 1
stage 4:  enumerations = 7
stage 3:  enumerations = 142
stage 2:  enumerations = 887
stage 1:  enumerations = 0

           |               col
       row |         1          2          3 |     Total
-----------+---------------------------------+----------
         1 |         9          2          8 |        19 
         2 |         2          3          2 |         7 
         3 |         1          6          2 |         9 
         4 |         5          3          5 |        13 
         5 |         3          6          3 |        12 
-----------+---------------------------------+----------
     Total |        20         20         20 |        60 

           Fisher's exact =                 0.167

Most careful accounts of the problem of small observed frequencies emphasise that the real danger zone is when they go below 1. An often overlooked example is Harold Jeffreys, Theory of Probability. 
A: The phrasing

any of the features of interest was more likely to occur than no feature of interest: 48 paragraphs with features, 12 features without, p < .001

suggests to me a test of the null hypothesis that the probability of a paragraph's having some feature of interest is one-half, using the binomial distribution. Why that null hypothesis should be of particular interest is another question.
A: 
What significance test might have provided the p < .001 finding?

Nick Cox has provided information about analysis of the table you provide using the $\chi^2$ test and an exact test. Neither resulted in p<0.001.
The results are presented in the paper as follows:

The only way I could think that the authors of the paper arrived at this p-value is to apply a test of proportion on the marginal result that 48/60 showed some features and 12/60 showed no features. See below, from Stata
. prtesti 60 48 60 12, count

Two-sample test of proportions                     x: Number of obs =       60
                                                   y: Number of obs =       60
------------------------------------------------------------------------------
    Variable |       Mean   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
           x |         .8   .0516398                      .6987879    .9012121
           y |         .2   .0516398                      .0987879    .3012121
-------------+----------------------------------------------------------------
        diff |         .6   .0730297                      .4568645    .7431355
             |  under Ho:   .0912871     6.57   0.000
------------------------------------------------------------------------------
        diff = prop(x) - prop(y)                                  z =   6.5727
    Ho: diff = 0

    Ha: diff < 0                 Ha: diff != 0                 Ha: diff > 0
 Pr(Z < z) = 1.0000         Pr(|Z| > |z|) = 0.0000          Pr(Z > z) = 0.0000

This is erroneous, of course, as pointed out by @whuber in the comments below.
Nevertheless, it is one way to explain the results they've reported.
