Let the average be denoted by $X$. We know that:
$X \sim N(\mu,\sigma^2)$
where
$\mu$ and $\sigma^2$ depend on the range over which you sample.
The problem then is how to transform $X$ to be uniform between 3 and 7. We can use the probability integral transform to accomplish this. Specifically, if we let:
$Y = F_X(X)$
where $F_X(.)$ is the cdf of $X$.
Then it follows that:
$Y \sim U(0,1)$
We can then rescale $Y$ to get the desired values to lie between 3 and 7 as follows:
$Z = 3 + 4 Y$
It is clear that $Z$ has the required property of being uniform between 3 and 7 as required.
Edit
The algorithm to follow would be the following:
Step 1: Generate 1000 random numbers uniformly between 0 and 10.
Step 2: Compute the average.
Step 3. Repeat steps 1 and 2 1000 times. So, now you have 1000 sample averages which should follow a normal distribution because of the central limit theorem.
Note: From statistical theory we know that the average values follow a normal distribution with mean 5 and variance $\frac{100}{12000}$
So. you could replace steps 1-3 above by drawing 1000 random variables from $N(5,\frac{100}{12000})$.
Step 4. For each one of the averages from step 3 (denote by X), compute $Y = F(X,5,\frac{100}{12000})$.
Here $F(X,5,\frac{100}{12000})$ is the cdf of the normal distribution centered at 5 and with variance $\frac{100}{12000}$
So, now we have 1000 values of $Y$ corresponding to the 1000 sample averages (i.e., $X$) from step 3. But, $Y$ is uniformly distributed between 0 and 1. Hence you need to transform each one of the values of $Y$ as outlined in step 5.
Step 5: Compute $Z=3+4Y$
Thus, you now have 1000 values (i.e., $Z$) which are distributed uniformly between 3 and 7 as desired.