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We will prove that the smallest sum is equal to $8$. One case when it is achieved is for $n=1$.

Suppose that for some $n>1$ it is possible to obtain a smaller sum than $8$. Observing the last digit of the number $5^n+6^n+2022^n$, we can easily conclude that $$5^n+6^n+2022^n\equiv \{\begin{array}{ll}7(\mathrm{mod}10),& \text{if }n\equiv 0(\mathrm{mod}4)\\ 3(\mathrm{mod}10),& \text{if }n\equiv 1(\mathrm{mod}4)\\ 5(\mathrm{mod}10),& \text{if }n\equiv 2(\mathrm{mod}4)\\ 9(\mathrm{mod}10),& \text{if }n\equiv 3(\mathrm{mod}4)\end{array}$$ It follows that $n\equiv 1,2(\mathrm{mod}4)$. We now consider these two cases.

Case 1: If $n=4k+1$, then $$5^n+6^n+2022^n\equiv \{\begin{array}{ll}5(\mathrm{mod}9),& \text{if }k\equiv 0(\mathrm{mod}3)\\ 2(\mathrm{mod}9),& \text{if }k\equiv 1(\mathrm{mod}3)\\ 8(\mathrm{mod}9),& \text{if }k\equiv 2(\mathrm{mod}3)\end{array}$$ From here, due to the last digit being equal to $3$, it is only possible for the sum of the digits to be equal to $5$. Since $5^n+6^n+2022^n\equiv 1(\mathrm{mod}4)$, the penultimate digit must be equal to $1$, and all other digits (except the first) are equal to $0$. So the last digits of $5^n+6^n+2022^n$ are $0013$ and therefore $5^n+6^n+2022^n\equiv 13(\mathrm{mod}16)$. On the other hand, since $n>1$ and $n\equiv 1(\mathrm{mod}4)$ we have $5^n+6^n+2022^n\equiv 5\cdot 5^{4k}\equiv 5(\mathrm{mod}16)$, a contradiction.

Case 2: If $n=4k+2$, then $$5^n+6^n+2022^n\equiv \{\begin{array}{ll}7(\mathrm{mod}9),& \text{if }k\equiv 0(\mathrm{mod}3)\\ 1(\mathrm{mod}9),& \text{if }k\equiv 1(\mathrm{mod}3)\\ 4(\mathrm{mod}9),& \text{if }k\equiv 2(\mathrm{mod}3)\end{array}$$ Due to the last digit, the only possibility is $k\equiv 0(\mathrm{mod}3)$ so $n=12s+2$ for some $s\in {\mathbb{N}}_0$. Now the sum of the digits is $7$, and the last digit is $5$. Let's use the divisibility criterion with $37$. (A number when divided by $37$ gives the same remainder as the sum of its three-digit blocks that are formed from right to left.) Since $6^n+5^n+2022^n$ must be equal to one of $20\cdots 05$ or $10\cdots 010\cdots 05$, according to the above criterion $6^n+5^n+2022^n\equiv 205,25,7,115,106,16(\mathrm{mod}37)$. I.e. congruent to $20,25,7,4,32,16(\mathrm{mod}37)$.

On the other hand, since it is easy to check that $6^{12}\equiv 1(\mathrm{mod}37)$, $5^{12}\equiv 10(\mathrm{mod}37)$ and $2022^{12}\equiv 24^{12}\equiv 10(\mathrm{mod}37)$ we have that $$5^n+6^n+2022^n\equiv 25\cdot 10^2+36+24^2\cdot 10^2\equiv 36+9\cdot 10^s\equiv \{\begin{array}{ll}8(\mathrm{mod}37),& \text{if }s\equiv 0(\mathrm{mod}3)\\ 15(\mathrm{mod}37),& \text{if }s\equiv 1(\mathrm{mod}3)\\ 11(\mathrm{mod}37),& \text{if }s\equiv 2(\mathrm{mod}3)\end{array}$$ This is a contradiction.

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Alternative solution.

Let $A_n=5^n+6^n+2022^n$. We have $A_n\equiv 1(\mathrm{mod}4)$. For $n\equiv 1,2,\dots ,5(\mathrm{mod}5)$ we have $6^n\equiv 6,11,16,21,1(\mathrm{mod}25)$ and for $n\equiv 1,2,\dots ,20(\mathrm{mod}20)$ we have $$2022^n\equiv -3,9,-2,6,7,4,13,11,17,-1,3,-9,2,-6,-7,-4,-13,-11,-17,1(\mathrm{mod}25)$$ Thus for $n>1$ and $n\equiv 1,2,\dots ,20(\mathrm{mod}20)$ we have $$A_n\equiv 3,20,14,27,8,10,24,27,38,0,9,2,18,15,-6,2,-2,5,4,2(\mathrm{mod}25)$$ So for $n>1$ and $n\equiv 1,2,\dots ,20(\mathrm{mod}20)$ we have $$A_n\equiv 53,45,89,77,33,85,49,77,13,25,9,77,93,65,69,77,73,5,29,77(\mathrm{mod}100)$$ The only possibilities for the sum of the digits of $A_n$ to be less than $8$ are $n\equiv 5,9,18(\mathrm{mod}20)$ where the last two digits of $A_n$ are $33,13,05$ respectively. (The case $n\equiv 10(\mathrm{mod}20)$ is rejected since then we must have $A_n=25$ which is impossible.)

Now for $n>1$ and $n\equiv 1,2,\dots ,6(\mathrm{mod}6)$ we have $A_n\equiv 5,7,8,4,2,1(\mathrm{mod}9)$. Looking modulo $60$, the only possibilities are $n\equiv 5,9,18,25,29,38,45,49,58(\mathrm{mod}60)$ and in those cases the last two digits of $A_n$ are $33,13,05,33,13,05,33,13,05$ respectively and the sum of digits of $A_n$ are $2,8,1,5,2,7,8,5,4$ $(\mathrm{mod}9)$ respectively.

So the only possibilities are $n\equiv 38,49(\mathrm{mod}60)$ where the last two digits of $A_n$ are $05$ and $13$ respectively, and the sum of digits of $A_n$ are $7$ and $5$ respectively.

The only possibilities are therefore for $A_n$ to be equal to a number of the form $20\dots 05$ or $10\dots 010\dots 05$ or $10\dots 013$. In particular, using the alternating sum of digits criterion for divisibility by $11$ we have that the first number is congruent to $3,7(\mathrm{mod}11)$, the second congruent to $3,5,7(\mathrm{mod}11)$ and the third congruent to $1,3(\mathrm{mod}11)$.

We now look at $A_n(\mathrm{mod}11)$. It can be easily checked that for $n\equiv 8(\mathrm{mod}10)$ we have $$A_n\equiv 4+4+3\equiv 0(\mathrm{mod}11)$$ and for $n\equiv 9(\mathrm{mod}10)$ we have $$A_n\equiv 9+2+5\equiv 5(\mathrm{mod}11)$$ So all three cases lead to a contradiction.