UPDATE: Seeing again the same ugly counting problem we can try a different approach using a ordinary generating function.
Let $s:=\{s_1,s_2,s_3,s_4\}$ the setup that we want to count where $s_k$ is the number of $k$-valued sides in the throw (by example if $s_3=5$ it means that the throw must contain exactly five $3's$.)
Because we are throwing $n$ dices then for each valid setup $s$ there are $\ell:=n-\sum_k s_k$ dice who values are $5$ or $6$. Then there is a maximum number $\ell$ of "wildcards" (that is, of possible $6's$). Let define the order relation $s\ge t$ if and only if $s_k\ge t_k$ for $k=1,2,3,4$. Then if $s\ge t$ then it is enough to have a valid throw when $\ell\ge s_6\ge\sum_k(s_k-t_k)=:|s-t|_1$.
Then for some $s\ge t$ there are $\ell+1-|s-t|_1$ possible valid setups. Now the ordinary generating function defined by
$$g_s(x):=\prod_{k=1}^4\left(\sum_{j=0}^{s_k}x^j\right)$$
is counting all $s\ge t$ for each $|t|_1$, that is $[x^{|t|_1}]g_s(x)$ is the number of $t\le s$ with $|t|_1:=\sum_k t_k$. And because for every $|t|_1$ there are $\ell+1-|s-t|_1=n+1-2|s|_1+|t|_1$ valid setups to count them all it is enough to make the dot product $(v_s,r_s)$, where $r_s$ is the vector defined by the coefficients of the polynomial $g_s$, that is
$$r_s:=([x^0]g_s(x),[x^1]g_s(x),\ldots,[x^{|s|_1}]g_s(x))$$
and $v_s:=(n+1-|s|_1,\ldots,2,1,0,\ldots,0)$.
Then the probability $p_s$ that some setup equivalent to $s$ appear in this game will be $p_s=(v_s,r_s)/6^n$. From this notation it could be easier to write efficient programs that evaluate these probabilities.
However if $n$ is too big probably an approximation could be better, by example from some Montecarlo simulation. In any case I doubt that a "closed" formula can be found.
This may help: the chance of getting a 1 is $\frac{1}{6}$, so the chance of getting seven 1s is $(\frac{1}{6})^7$.
However, we also want to count getting seven 2s, seven 3s, etc. There are six numbers, so now we have a chance of $6\cdot(\frac{1}{6})^7$ which is $(\frac{1}{6})^6$
Best Answer
The probability of getting exactly five ones and the remaining five as other different numbers would be
$$\binom{10}{5}\left(\frac{1}{6}\right)^5\left(\frac{5}{6}\right)^5$$
Multiplying by $6$ would almost have given us the probability of having some number that we have exactly five of, however it overcounts the cases where we have two numbers, each of which occurring five times.
The probability then that we have some number that we have exactly five of us then $$6\binom{10}{5}\left(\frac16\right)^5\left(\frac56\right)^5-\binom{6}{2}\binom{10}{5}\left(\frac16\right)^{10}$$
If we are interested also in the cases where we have some number occurring more than five times, we would also want to add $6\binom{10}{6}\left(\frac{1}{6}\right)^6\left(\frac{5}{6}\right)^4,6\binom{10}{7}\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)^3,\dots$ to account for the cases of having exactly six, exactly seven, etc... of a number appearing (noting that it is impossible to have strictly more than five of one number and five or more of another number simultaneously).
Combining the above into a more compact form, we have as a final probability:
$$\sum\limits_{k=5}^{10}\left[6\binom{10}{k}\left(\frac16\right)^k\left(\frac56\right)^{10-k}\right]-\binom{6}{2}\binom{10}{5}\left(\frac16\right)^{10}$$
Related reading: Binomial Distribution on Wikipedia.