Do not histogram with an equal number of bins: histogram with consistent edges to the bins. Then you can just put all the results together and mean() as appropriate.
In the below example, numedges-1 and edges(1:end-1) relates to the fact that the number of bins produced by histcounts is one fewer than the number of edges, because the final edge marks the end of the range.
num_datasets = 100;
edges = sort(rand(1, 14) * 5);
numedges = length(edges);
counts = zeros(numedges-1, num_datasets);
for dataset_idx = 1 : 100
counts(:, dataset_idx) = histcounts( datasets{dataset_idx}, edges );
end
avg_counts = mean(counts);
bar(edges(1:end-1), avg_counts)
