Considering that sub lists can have different size, get the minimum size or depending on need make it max

function size(ar){
    var row_count = ar.length;
    var row_sizes = []
    for(var i=0;i<row_count;i++){
        row_sizes.push(ar[i].length)
    }
    return [row_count, Math.min.apply(null, row_sizes)]
}
size([[1, 1, 1], [1, 1, 1]])

Output:

[2, 3]

Here is a simple recursive function to do that. Wrote it very quickly, so may prone to errors.

This function returns the lengths of each dimensions. If it contains sub-arrays of different lengths, this function returns false.

This function works for different dimensions.

For example, passing a 2d array of 2x3 will return [2, 3], a 3d array of 2x3x4 will return [2, 3, 4]. The dimension (i.e. n d) can be obtained by checking the length of the returned array.

// Array dimension checker
// Returns:
//   false when array dimensions are different
//   an Array when is rectangular 0d (i.e. an object) or >=1d
function arrayDimension(a) {
    // Make sure it is an array
    if (a instanceof Array) {
        // First element is an array
        var sublength = arrayDimension(a[0]);
        if (sublength === false) {
            // Dimension is different
            return false;
        } else {
            // Compare every element to make sure they are of the same dimensions
            for (var i = 1; i < a.length; i++) {
                var _sublength = arrayDimension(a[i]);
                // HACK: compare arrays...
                if (_sublength === false || sublength.join(",") != _sublength.join(",")) {
                    // If the dimension is different (i.e. not rectangular)
                    return false;
                }
            }
            // OK now it is "rectangular" (could you call 3d "rectangular"?)
            return [a.length].concat(sublength);
        }
    } else {
        // Not an array
        return [];
    }
}

Demo (Check the javaScript console :) )