Of the $6^{10}$ possible rolls, $6^{10}-5^{10}-10\cdot5^9$ have two $1$’s. Of the $10\cdot5^9$ rolls that have exactly one $1$, $10(5^9-4^9)$ have a $2$ as the lowest of the remaining nine dice; $10(4^9-3^9)$ have a $3$; $10(3^9-2^9)$ have a $4$; $10(2^9-1)$ have a $5$; and $10$ have a $6$.
Of the $5^{10}$ rolls that have no $1$’s, $5^{10}-4^{10}-10\cdot4^9$ have two $2$’s. Of the $10\cdot4^9$ rolls that have exactly one $2$, $10(4^9-3^9)$ have a $3$ as the lowest of the remaining nine dice; $10(3^9-2^9)$ have a $4$; $10(2^9-1)$ have a $5$; and $10$ have a $6$.
Of the $4^{10}$ rolls that have no $1$’s or $2$’s, $4^{10}-3^{10}-10\cdot3^9$ have two $3$’s. Of the $10\cdot3^9$ rolls that have exactly one $3$, $10(3^9-2^9)$ have a $4$ as the lowest of the remaining nine dice; $10(2^9-1)$ have a $5$; and $10$ have a $6$.
Of the $3^{10}$ rolls that have no $1$’s, $2$’s, or $3$’s, $3^{10}-2^{10}-10\cdot2^9$ have two $4$’s. Of the $10\cdot2^9$ rolls that have exactly one $4$, $10(2^9-1)$ have a $5$ as well, and $10$ have only $6$’s.
Of the $2^{10}$ rolls that have only $5$’s and $6$’s, $2^{10}-1-10$ have two $5$’s, and $10$ have a $5$ and nine $6$’s.
There is one roll whose smallest two dice are both $6$’s.
The sum of the numbers on the removed dice is therefore
$\begin{align*} &2\left(6^{10}-5^{10}-10\cdot5^9\right)+10\sum_{k=2}^6\left((7-k)^9-(6-k)^9\right)(k+1)\\ &\qquad+4\left(5^{10}-4^{10}-10\cdot4^9\right)+10\sum_{k=3}^6\left((7-k)^9-(6-k)^9\right)(k+2)\\ &\qquad+6\left(4^{10}-3^{10}-10\cdot3^9\right)+10\sum_{k=4}^6\left((7-k)^9-(6-k)^9\right)(k+3)\\ &\qquad+8\left(3^{10}-2^{10}-10\cdot2^9\right)+10\sum_{k=5}^6\left((7-k)^9-(6-k)^9\right)(k+4)\\ &\qquad+10\left(2^{10}-1-10\right)+10\cdot11\\ &\qquad+12\\ &=2\sum_{k=1}^6k^{10}+10\cdot5^9+20\cdot4^9+30\cdot3^9+40\cdot2^9+50\\ &=168,066,052\;, \end{align*}$
if I’ve made no computational errors.
The sum of all the dice in all $6^{10}$ possible rolls is $10\cdot3.5\cdot6^{10}=2,116,316,160$; subtracting the total of the two lowest numbers in each leaves $1,948,250,108$, for an average total per roll of $10$ dice of $32.22049477711$ and an average per die of $4.027561847139$. Henry’s simulation is right on the money in this case.