JavaScript (Node.js),  130  99 bytes

f=(n,p=`
`,q=`  `,k=` \\`)=>n?f(n>>1,p+(n&1?` `:`  `),n&1?k:`/\\`)+p+(n&1?q+k:` ${k+k+p+q+k}/`):p+q

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Pure bash 170

Reading other answers here and re-reading request, there is a version without integer to binary conversion:

l=${1//1};t=${#1};printf -vk '%*s' $[t+${#l}-1];p=;for((c=t;c--;)){
((${1:c:1}))&&echo "$k\ $p"&&p=\\||{
k=${k:1};echo "$k/\ $p"$'\n'"$k\ \\";p=\\/;};k=${k:1};};echo \ \\

Pure bash 204

Full version expecting integer as argument.

for((a=$1;a;a>>=1)){ s=$[a&1]$s;};l=${s//1} t=${#s};printf -vk '%*s' $[t+${#l}-1]
p=;for((c=t;c--;)){ ((${s:c:1}))&&echo "$k\ $p"&&p=\\||{
k=${k:1};echo "$k/\ $p"$'\n'"$k\ \\";p=\\/;};k=${k:1};};echo \ \\

GolfScript, 82 81 76 75 bytes

{49='/\ \  \ \/ \ \ '10/=}%5/zip.3<{' ':a+}%\3>{a\+}%+zip.,.a*:b@*b\+\6+/n*

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LSB at the bottom left.

Python 3.8 (pre-release), 120 118 bytes

def f(i):
 c=a,=["\n"]
 while i:c[0]+=" \\"+(x:=~i%2)*"/";c=[a+x*"/"+"\\",x*(a+"\\ \\")]+c;a+=-~x*" ";i//=2
 print(*c)

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Output with a leading newline, some trailing whitespaces at the end of each line and with the least significant bit in the bottom left (LSB)

Zsh, 88 bytes

for b;((b))&&L=(\ $^L \\$l)&&l=' \'||{L=('  '$^L '/\'$l '\ \');l=' \/' }
print -rl $L $l

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Or add --rcexpandparam to remove the ^ for 86 bytes.

for bit;
    ((bit)) &&
        # prepend all but one of the previous lines with a space,
        # the last line prefixed with a '\'
        lines=(\ $^lines \\$last) &&
        # new last line is ' \'
        last=' \' ||
    {
        
        # prepend all but one of the previous lines with two spaces,
        # the last line prefixed with a '/\',
        # add '\ \' as the center of the zero
        lines=('  '$^lines '/\'$last '\ \')
        # new last line is ' \/'
        last=' \/'
    }
print -rl $lines $last

JavaScript (ES6), 102 bytes

Expects an integer and outputs with the LSB at the bottom left.

f=(n,i,p=`
`,b=n-~n>>i/2)=>b>1?f(n,b/2%2-~i,p+" ")+p+"/ "[b&1|i&1]+s[+!!i]+s+"/  "[b&2|i&1]:p+=s=" \\"

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Commented

f = (          // f is a recursive function taking:
  n,           //   n = input integer
  i,           //   i = counter (initially undefined)
  p = `\n`,    //   p = prefix string
  b = n - ~n   //   b = n * 2 + 1 right shifted by floor(i / 2)
      >> i / 2 //
) =>           //
b > 1 ?        // if b is greater than 1:
  f(           //   do a recursive call:
    n,         //     pass n unchanged
    b / 2 % 2  //     add 2 to i if the bit #1 of b is set
    - ~i,      //     or 1 otherwise (*)
               //     (because zeros are twice as large as ones)
    p + " "    //     append a space to the prefix string p
  ) +          //   end of recursive call
  p +          //   append the prefix string p
  "/ "[        //   append a space if:
    b & 1 |    //     the target digit is a one
    i & 1      //     or i is odd (grid misalignment)
  ] +          //   or a "/" otherwise (the top of a zero)
  s[+!!i] +    //   append "\" if i > 0, or a space otherwise
               //   (top segment of previous digit)
  s +          //   append " \" (bottom segment of new digit)
  "/  "[       //   append a space if:
    b & 2 |    //     the target digit is a one
    i & 1      //     or i is odd (grid misalignment)
  ]            //   or a "/" otherwise (the bottom of a zero)
:              // else (end of recursion):
  p +=         //   append p
    s = " \\"  //   followed by " \" (top segment of MSB)

Perl 5 -F, 104 97 bytes

@;=((($b=$"x(length($;[0].=" \\".'/'x!$_)-3+$_))."/\\","$b\\".' \\'x!$_)[$_..1],@;)for@F;say for@

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APL+WIN, 66 bytes

Prompts for bit vector. Index origin = 0

n←(⊂3 2⍴'  \  \'),⊂3 3⍴'/\ \ \ \'⋄(-⌽⍳⍴s)⊖(i,i)⍴((i←⍴s)*2)↑n[s←~⎕]

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Python 3.8 (pre-release),  263   243   229   226   225   187   168   151  150 bytes

-20 bytes by applying some small modifications.
-14 bytes by emanresu A.
-3 bytes by noodle person.
-75 bytes by ASCII-only.
-1 byte by noodle person.

…Okay, what—

def f(b,c=[]):
 for i in b:c+=[0,1,2][i::2]
 for i in range(l:=len(c),-1,-1):print(" "*i+"  /"[[0,*c][i]]+" \\"[i>0]+" \\"*(i<l)+"  /"[[*c,0,0][i+1]])

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Charcoal, 23 17 bytes

ＦＳ¿Σι↗\«\↑/↗\»‖Ｍ↖

Try it online! Link is to verbose version of code. Takes input as a binary string. Explanation:

ＦＳ

Loop over each digit.

¿Σι

If it's a 1, then...

↗\

... output the bottom half of the 1, finishing with the cursor moved up and right.

«\↑/↗\»

Otherwise, draw the bottom half of the zero, finishing with the cursor moved up and right.

‖Ｍ↖

Reflect to complete the output.

Alternative solution, also 17 bytes:

ＦＳ«Ｆ¬Σι↑\¶/↗\»‖Ｍ↖

Try it online! Link is to verbose version of code. Explanation:

ＦＳ«

Loop over each digit.

Ｆ¬Σι

If it's a zero, then...

↑\¶/

... draw the \/ part.

↗\

Draw the bottom half of the 1 or complete the bottom half of the zero, finishing with the cursor moved up and right.

»‖Ｍ↖

Reflect to complete the output.

Edit: Saved 6 bytes by using ReflectMirror(:UpLeft); as suggested by @noodleman.

J, 78 bytes

' \/'{~[:+&(+|:&.|.)/1 2([*Q=/~([+|.@Q=:i.@#@])*_ 1{~]>.~2-[)"{0,&;(0 1;0){~|.

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Accepts input as a list of binary digits. Here's the expanded version I started with, which shows some of the symmetry:

b =: 1 1 0 1 0 1
mask =: 0,;(0 1;0){~|.b

e =: [ + i.@-@#@]
mK =: 2 e mask
mT =: 1 e mask

K =: 2 * (mK *_ 1{~mask) =/ i.#mask
T =: 1 * (mT           ) =/ i.#mask

echo ' \/' {~ (+|:&.|.) K+T

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