Create a turing machine for log base 2 of n
The value of $\lceil\log_2(n)\rceil$ is more or less equal to the number of bits in the binary representation of $n$ (except if $n$ is a power of 2).
You can then imagine a Turing Machine with an input tape (on which is written the number), a work tape and an output tape. First, you need to convert your unary number on the input tape, into binary on the work tape. This can be done with a simple incrementation Turing Machine (you can decide on the side of the least significant bit, but imho left is easier). Then, you need to count the number of bits in the binary number, and write a $1$ for each bit counted. You can use an additionnal state, to detect is the number is a power of 2 (a $1$ followed only by $0$).
This machine can be reduced to two tapes by replacing bits by $1$ on the work tape (except maybe for the last one which needs to become $\#$ if $n$ is a power of $2$). It can be further reduced to one tape by using the read part of the tape to write the binary number (and another symbol than $\#$ to replace the $1$ which are read).
Let's be more formal:
I consider a Turing Machine $(Q, \Sigma, \Gamma, \delta, q_0, q_f, \#)$ where $Q$ is the set of states of the TM; $\Sigma = \{1\}$ is the input alphabet; $\Gamma = \{0, 1, \\\$, \#\}$ is the work alphabet ($\\\$$ is a symbol to separate the input and the output), $\delta$ is the transition function, $q_0\in Q$ is the initial state, $q_f \in Q$ is the final state and $\#$ is the blank symbol. The Turing Machine I imagined is composed of 13 states, it may be possible to do better. Here is the detail on each state (to understand how the TM works) :

$q_0$: initial state
$q_1$: I have read only one $1$ in the input
$q_2$: I have read only two $1$'s in the input
$q_3$: I am reading another $1$ in the input ($q_1$ and $q_2$ make it easier to separate the input and the output)
$q_4$: I am returning to the most left to increment the binary representation
$q_5$: I am incrementing, and there is still a carry
$q_6$: I have finished incrementing, I return to the input to read $1$'s
$q_7$: I have finished reading completely the input, I am returning to the most left to transform the binary representation of $n$ into unary representation of $\lceil\log_2 n\rceil$
$q_8$: I am replacing $0$'s by $1$'s without having seen a single $1$
$q_9$: I am replacing $0$'s by $1$'s having seen a single $1$
$q_{10}$: I am replacing $0$'s by $1$'s having seen at least two $1$'s (therefore $n$ is not a power of $2$)
$q_{11}$: I have concluded that $n$ is a power of $2$, so the last $1$ must be replaced by $\#$
$q_f$: final state

Here is the detail of the transition function, state by state. The first column is the current state of the Turing Machine, the last 4 correspond to the read tape symbol. Each box is given as a triple $(\alpha, q, d)$ where $\alpha$ is the symbol written on the tape, $q$ is the next state and $d$ is the direction of the head shift.




State
$0$
$1$
$\\\$$
$\#$




$q_0$
-
$1, q_1, \rightarrow$
-
$\#, q_f, \rightarrow$


$q_1$
-
$1, q_2, \rightarrow$
-
$\#, q_{11}, \leftarrow$


$q_2$
-
$\\\$, q_3, \rightarrow$
-
$\#, q_{11}, \leftarrow$


$q_3$
-
$\\\$, q_4, \leftarrow$
$\\\$, q_3, \rightarrow$
$\#, q_{7}, \leftarrow$


$q_4$
$0, q_4, \leftarrow$
$1, q_4, \leftarrow$
$\\\$, q_4, \leftarrow$
$\#, q_{5}, \rightarrow$


$q_5$
$1, q_6, \rightarrow$
$0, q_5, \rightarrow$
$1, q_6, \rightarrow$
-


$q_6$
$0, q_6, \rightarrow$
$1, q_6, \rightarrow$
$\\\$, q_3, \rightarrow$
-


$q_7$
$0, q_7, \leftarrow$
$1, q_7, \leftarrow$
$\#, q_7, \leftarrow$
$\#, q_8, \rightarrow$


$q_8$
$1, q_8, \rightarrow$
$1, q_9, \rightarrow$
-
-


$q_9$
$1, q_9, \rightarrow$
$1, q_{10}, \rightarrow$
-
$\#,q_{11},\leftarrow$


$q_{10}$
$1, q_{10}, \rightarrow$
$1, q_{10}, \rightarrow$
-
$\#,q_f,\leftarrow$


$q_{11}$
-
$\#, q_f, \rightarrow$
-
-




Here is a simulation of this TM : http://turingmachinesimulator.com/shared/eikrrbjoiz