It does not appear that a standard definition of AIC is used in FindDistribution. The usual formula with $n$ observations and $k$ parameters is $AIC=-2 \log L+2 k$. But FindDistribution seems to use $AIC=2 \log L - 2k/(n-k-1)$.

Here's "proof" of what FindDistribution uses. (This works unless a MixtureDistribution is found with more than two distributions.)

Set the sample size and get a random sample:

n = 100;
data = RandomVariate[Exponential[1], n];

Find the top best fitting distributions and collect the AIC and LogLikelihood values:

nbest = 5;
fd = FindDistribution[data, nbest, {"AIC", "LogLikelihood"}]
(* {{ExponentialDistribution[1.07497], {-1.83436, -0.906976}},
{MixtureDistribution[{0.72248, 0.27752}, {GammaDistribution[1.2916, 0.369832], UniformDistribution[{0.0177466, 3.77159}]}], {-1.75598, -0.824797}},
{MixtureDistribution[{0.765586, 0.234414}, {GammaDistribution[1.3553, 0.342329], NormalDistribution[2.45319, 0.808439]}],{-1.78878, -0.841199}},
{LogNormalDistribution[-0.708397,1.24473], {-1.84787, -0.903319}},
{WeibullDistribution[0.92819,0.897149], {-1.85254, -0.905651}}} *)

Find $k$ (the number of parameters):

k = StringCount[Table[ToString[fd[[i, 1]]], {i, nbest}], ","] + 1;
(* {1,6,6,2,2} *)
nMixtures = StringCount[Table[ToString[fd[[i, 1]]], {i, nbest}], "MixtureDistribution"]
(* {0,1,1,0,0} *)
k = k - nMixtures
(* {1,5,5,2,2} *)

Extract the LogLikelihood and AIC values and show the AIC values from the formula used by FindDistribution:

logL = Table[fd[[i, 2, 2]], {i, nbest}]
(* {-0.9069763208709088`,-0.8247971259593575`,-0.8411993739988386`,-0.903318900970027`,-0.9056506499137531`} *)
aic = Table[fd[[i, 2, 1]], {i, nbest}]
(* {-1.8343608050071236`,-1.7559772306421193`,-1.7887817267210813`,-1.847874915342116`,-1.8525384132295681`} *)
2 logL - 2 k/(n - k - 1)
(* {-1.8343608050071238`,-1.7559772306421193`,-1.7887817267210815`,-1.847874915342116`,-1.8525384132295681`} *)

It appears that the formula used is wrong and that it seems to be a combination of $AIC$ and $AIC_c$ as $AIC_c =-2\log L + 2 k n/(n-k-1)$.

I was too mild in my assessment: the AIC definition used by FindDistribution is wrong. So I'm a assuming that it is a coding error rather than an alternative formulation of AIC. (Likely the intent was to use $AIC_c$ but the sample size ( $n$) in the numerator for the number of parameters adjustment was left out of the equation. But someone will correct me if I'm wrong.) That wrong definition way under-corrects for the number of parameters in a model so the ranking of models from FindDistribution is suspect (if the number of parameters vary among the models considered).

To make the usual statements about $AIC$: this measure allows the ranking of models and cannot be used as a measure of absolute fit. The model with the best $AIC$ might be the best model of a bunch of bad models or the best of a bunch of very good models.

While FindDistribution can get you a parsimonious description of the distribution of your data (a few parameters + a particular family of distributions) that you can easily use outside of Mathematica, if you want a more justifiable description of the data, I'd suggest using SmoothKernelDistribution. You don't get a parsimonious description of your data but it will almost certainly provide a better fit.

Similarly, $AIC$ from LinearModelFit appears to be wrongly computed.

Did this ever get fixed or otherwise resolved? Having some issues about the AIC using LinearModelFit, but it could well be user error rather than a bug.

Not resolved as far as I know. And there is another issue I should have noticed before: the log likelihood is wrong also. The "mean" of minus the log likelihood is used (i.e., the log of the likelihood for the data and proposed model is divided by the sample size). The ranking of models isn't affected but if one attempts to apply the common threshold of "2 AIC units" suggesting a very different model, well, that won't work with the AIC values produced by FindDistribution.

I'll write up a summary and send to Wolfram, Inc., and report back. Below working example:

(* Generate data and find the 5 best fitting distributions *)
n = 100; (* Sample size *) 
SeedRandom[12345];
data = RandomVariate[ExponentialDistribution[1], n];
nbest = 5;
TableForm[(fd = FindDistribution[data, nbest, {"LogLikelihood", "AIC"}]),
 TableHeadings -> {("Distribution " <> # &) /@ (ToString[#] & /@ Range[nbest]), {"Distribution", "log(L) and AIC"}}]

(* Log of the likelihood as calculated by FindDistribution *)
Column[{Style["Log of the likelihood\n", 18, Bold],
  TableForm[Table[{fd[[i, 2, 1]], LogLikelihood[fd[[i, 1]], data]/n,
     LogLikelihood[fd[[i, 1]], data]}, {i, nbest}],
   TableHeadings -> {("Distribution " <> # &) /@ (ToString[#] & /@ Range[nbest]),
     {"\nFrom\nFindDistribution", "Duplicating what\nFindDistribution\ndoes",
      "What the\nlog likelihood\nshould be"}}]}]

(* AIC as calculated by FindDistribution *)
Column[{Style["AIC\n", 18, Bold], TableForm[Table[{fd[[i, 2, 2]],
     2 LogLikelihood[fd[[i, 1]], data]/n - 2 k[[i]]/(n - k[[i]] - 1),
     -2 LogLikelihood[fd[[i, 1]], data] + 2 k[[i]]}, {i, nbest}],
   TableHeadings -> {("Distribution " <> # &) /@ (ToString[#] & /@ Range[nbest]),
     {"\nFrom\nFindDistribution", "Duplicating what\nFindDistribution\ndoes",
      "\nWhat AIC\nshould be"}}]}]

(Note the sign of AIC is arbitrary. A common approach is to choose the "smaller is better" approach which is what I used for the "What AIC should be" column.)

Seth: You might want to post a specific question about AIC with LinearModelFit (and NonlinearModelFit) as the issue might not be user error or an error in Mathematica but rather a difference in definitions.

To use a slightly modified example from the documentation (I added an additional sample point) here is what LinearModelFit puts out:

data = {{0, 1}, {1, 0}, {3, 2}, {5, 4}, {7, 5}};
n = 5;  (* Sample size *)
p = 3;  (* Number of parameters: intercept, slope, and residual error variance *)
lm = LinearModelFit[data, x, x];
lm["AIC"]
(* 15.1332 *)

And here is what LinearModelFit is doing (in a slightly less-black box way):

aic = -2 LogLikelihood[NormalDistribution[0, lm["EstimatedVariance"]^0.5], lm["FitResiduals"]] + 2*p
(* 15.1332 *)

The issue is: Is the definition of AIC that LinearModelFit uses the definition that you want to use?

In essence restricted maximum likelihood (REML) is being performed (i.e., obtaining the unbiased estimate of variance) but rather than plugging in that estimate of variance into the log of the restricted likelihood function, that estimate is being plugged into the unrestricted log likelihood function. That is not what SAS or R uses to construct AIC. At best how LinearModelFit and NonlinearModelFit calculates AIC is not common practice.