Three possible solutions:
1) You can do a decent job simulating the average of 10 rolls in a single cell by using a normal approximation. In the target cell enter:
=ROUND(NORM.INV(RAND(),3.5,SQRT(35/(12*10))),1)
I left the 10 in explicitly to show what you would need to modify if you want to vary the number of rolls. The fact that 35 is 3.5 * 10 is a coincidence, the number 35 would stay there even with 100 rolls. If you vary from 10, you might want to tweak the rounding. For 10 the standard deviation of is roughly 0.54, putting 1 and 6 at around 4.6 standard deviations away from the mean of 3.5. On average the formula will give you a value outside the range 1-6 1 out of every 270,000 times. A histogram of dice averages for 10 rolls forms a fairly nice bell-shaped curve, so this approach is certainly reasonable, and might work for your purposes.
2) A VBA sub which fills selected cells with exact spreadsheet formulas, relieving you of the tedious need to count. If there is a requirement that the final spreadsheet is macro-free, this can still be used in the development stage:
Sub DiceAverageFormula()
'places a formula in the selected cell (or cells)
'which simulates the average of rolling a die n 10 times
Dim i As Long, n As Long
Dim form As String
n = InputBox("How many times do you want to roll?", "Dice", 10)
If n < 1 Then Exit Sub
form = "=Average(Randbetween(1,6)"
For i = 2 To n
form = form & ",Randbetween(1,6)"
Next i
form = form & ")"
Selection.Formula = form
End Sub
3) For completeness sake, The naïve VBA UDF:
Function RollDice(n As Long, Optional k As Long = 6) As Double
'simulates the rolling of n k-sided die, returing the average
Application.Volatile 'can be removed, of course
Dim i As Long, sum As Long
With Application.WorksheetFunction
For i = 1 To n
sum = sum + .RandBetween(1, k)
Next i
End With
RollDice = sum / n
End Function
To test the various approaches, I filled 30 rows with each approach and made 5-number summaries of the results (which appear reasonable):

Needless to say, the fact that in this screenshot the mean for the normal approximation is lower than the other two is not particularly relevant. You would need a fairly large number of runs before a statistical test could detect any difference between the first column and the other 2.
On Edit: Here is a graph of exact probability vs. normal approximation for the mean of 10 rolls of a fair die. It underscores how reasonable it is to use the central limit theorem here. Even though n = 10 is a somewhat small sample size, the exact probability is sufficiently symmetric and unimodal that it already looks fairly normal:
