$\uparrow$ More Up-Arrows $\uparrow$

Let $p(m,n)=m^n \mod 10$

It is clear that for all $n>0$ , $p(1,n)=1,p(5,n)=5,p(6,n)=6$

Further, for $m=2,4,8$ , we have $m^2 \mod 4 = 0$ and thus $m\uparrow\uparrow\uparrow (n+1) = m^p$ where $p$ would be a multiple of 4 (because m is a even number).

Thus, for even m, $m\uparrow\uparrow\uparrow (m+1)$ would be of the form $m^{4n}$

$p(2,4n)=6,p(4,4n)=6,p(8,4n)=6$

For $m=3,7$ , let

$m\uparrow\uparrow\uparrow (m+1) = m^{m^q}$

Now, $m^q \mod 4 = (m \mod 4)^q = (-1)^q = (-1)$ (because $q$ would be an odd number for $m=3,7$ and $3 \mod 4 = 7 \mod 4 = -1 \mod 4 = 3$ .

Hence, $3\uparrow\uparrow\uparrow 4 \mod 10 = 3^3 \mod 10 = 7$ and $7\uparrow\uparrow\uparrow 8 \mod 10 = 7^3 \mod 10 =3$

Finally, it can be seen that any odd power of $9$ would end in $9$ , hence $9\uparrow\uparrow\uparrow 10 \mod 10 = 9$

Hence, the required result is

$1+6+7+6+5+6+3+6+9=\boxed{49}$